Count the coins

From Rosetta Code
(Redirected from Count the Coins)
Task
Count the coins
You are encouraged to solve this task according to the task description, using any language you may know.

There are four types of common coins in   US   currency:

  1.   quarters   (25 cents)
  2.   dimes   (10 cents)
  3.   nickels   (5 cents),   and
  4.   pennies   (1 cent)


There are six ways to make change for 15 cents:

  1.   A dime and a nickel
  2.   A dime and 5 pennies
  3.   3 nickels
  4.   2 nickels and 5 pennies
  5.   A nickel and 10 pennies
  6.   15 pennies


Task

How many ways are there to make change for a dollar using these common coins?     (1 dollar = 100 cents).


Optional

Less common are dollar coins (100 cents);   and very rare are half dollars (50 cents).   With the addition of these two coins, how many ways are there to make change for $1000?

(Note:   the answer is larger than   232).


References



11l

Translation of: Python
F changes(amount, coins)
   V ways = [Int64(0)] * (amount + 1)
   ways[0] = 1
   L(coin) coins
      L(j) coin .. amount
         ways[j] += ways[j - coin]
   R ways[amount]

print(changes(100, [1, 5, 10, 25]))
print(changes(100000, [1, 5, 10, 25, 50, 100]))

Output:

242
13398445413854501

360 Assembly

Translation of: AWK
*        count the coins           04/09/2015
COINS    CSECT
         USING  COINS,R12
         LR     R12,R15
         L      R8,AMOUNT          npenny=amount
         L      R4,AMOUNT
         SRDA   R4,32
         D      R4,=F'5'
         LR     R9,R5              nnickle=amount/5
         L      R4,AMOUNT
         SRDA   R4,32
         D      R4,=F'10'
         LR     R10,R5             ndime=amount/10
         L      R4,AMOUNT
         SRDA   R4,32
         D      R4,=F'25'
         LR     R11,R5             nquarter=amount/25
         SR     R1,R1              count=0
         SR     R4,R4              p=0
LOOPP    CR     R4,R8              do p=0 to npenny
         BH     ELOOPP
         SR     R5,R5              n=0
LOOPN    CR     R5,R9              do n=0 to nnickle
         BH     ELOOPN
         SR     R6,R6
LOOPD    CR     R6,R10             do d=0 to ndime
         BH     ELOOPD
         SR     R7,R7              q=0
LOOPQ    CR     R7,R11             do q=0 to nquarter
         BH     ELOOPQ
         LR     R3,R5              n
         MH     R3,=H'5'
         LR     R2,R4              p
         AR     R2,R3
         LR     R3,R6              d
         MH     R3,=H'10'
         AR     R2,R3
         LR     R3,R7              q
         MH     R3,=H'25'
         AR     R2,R3              s=p+n*5+d*10+q*25
         C      R2,=F'100'         if s=100
         BNE    NOTOK
         LA     R1,1(R1)           count=count+1
NOTOK    LA     R7,1(R7)           q=q+1
         B      LOOPQ
ELOOPQ   LA     R6,1(R6)           d=d+1
         B      LOOPD
ELOOPD   LA     R5,1(R5)           n=n+1
         B      LOOPN
ELOOPN   LA     R4,1(R4)           p=p+1
         B      LOOPP
ELOOPP   XDECO  R1,PG+0            edit count
         XPRNT  PG,12              print count
         XR     R15,R15
         BR     R14
AMOUNT   DC     F'100'             start value in cents
PG       DS     CL12
         YREGS
         END    COINS
Output:
         242

Ada

Works with: gnat/gcc
with Ada.Text_IO;

procedure Count_The_Coins is

   type Counter_Type is range 0 .. 2**63-1; -- works with gnat
   type Coin_List is array(Positive range <>) of Positive;

   function Count(Goal: Natural; Coins: Coin_List) return Counter_Type is
      Cnt: array(0 .. Goal) of Counter_Type := (0 => 1, others => 0);
      -- 0 => we already know one way to choose (no) coins that sum up to zero
      -- 1 .. Goal => we do not (yet) other ways to choose coins
   begin
      for C in Coins'Range loop
         for Amount in 1 .. Cnt'Last loop
            if Coins(C) <= Amount then
               Cnt(Amount) := Cnt(Amount) + Cnt(Amount-Coins(C));
               -- Amount-Coins(C) plus Coins(C) sums up to Amount;
            end if;
         end loop;
      end loop;
      return Cnt(Goal);
   end Count;

   procedure Print(C: Counter_Type) is
   begin
      Ada.Text_IO.Put_Line(Counter_Type'Image(C));
   end Print;

begin
   Print(Count(   1_00,          (25, 10, 5, 1)));
   Print(Count(1000_00, (100, 50, 25, 10, 5, 1)));
end Count_The_Coins;

Output:

 242
 13398445413854501

Alternate method that keeps track of the specific combinations of coins:

with Ada.Text_IO; use Ada.Text_IO;

procedure Main is
   count: Integer;
begin
   count := 0;
   for penny in 0 .. 100 loop
      for nickel in 0 .. 20 loop
        for dime in 0 .. 10 loop
           for quarter in 0 .. 4 loop
              if (penny + 5 * nickel + 10 * dime + 25 * quarter = 100)
              then
                 Put_Line(Integer'Image(count+1) & ": " &
                          Integer'Image(penny)   & " pennies, " &
                          Integer'Image(nickel)  & " nickels, " &
                          Integer'Image(dime)    & " dimes, "   &
                          Integer'Image(quarter) & " quarters");
                 count := count + 1;
              end if;
           end loop;
        end loop;
      end loop;
   end loop;
   
  Put_Line("The number of ways to make change for a dollar is: " &  Integer'Image(count));
end Main;

Output:

 1:  0 pennies,  0 nickels,  0 dimes,  4 quarters
 2:  0 pennies,  0 nickels,  5 dimes,  2 quarters
 3:  0 pennies,  0 nickels,  10 dimes,  0 quarters
 4:  0 pennies,  1 nickels,  2 dimes,  3 quarters
 5:  0 pennies,  1 nickels,  7 dimes,  1 quarters
.....................
 239:  90 pennies,  0 nickels,  1 dimes,  0 quarters
 240:  90 pennies,  2 nickels,  0 dimes,  0 quarters
 241:  95 pennies,  1 nickels,  0 dimes,  0 quarters
 242:  100 pennies,  0 nickels,  0 dimes,  0 quarters
The number of ways to make change for a dollar is:  242

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.4.1
Translation of: Haskell
#
  Rosetta Code "Count the coins"
  This is a direct translation of the "naive" Haskell version, using an array
  rather than a list. LWB, UPB, and array slicing makes the mapping very simple:
 
  LWB > UPB     <=> []
  LWB = UPB     <=> [x]
  a[LWB a]      <=> head xs
  a[LWB a + 1:] <=> tail xs
#

BEGIN
  PROC ways to make change = ([] INT denoms, INT amount) INT :
  BEGIN
    IF amount = 0 THEN
      1
    ELIF LWB denoms > UPB denoms THEN
      0
    ELIF LWB denoms = UPB denoms THEN
      (amount MOD denoms[LWB denoms] = 0 | 1 | 0)
    ELSE
      INT sum := 0;
      FOR i FROM 0 BY denoms[LWB denoms] TO amount DO
        sum +:= ways to make change(denoms[LWB denoms + 1:], amount - i)
      OD;
      sum
    FI
  END;
  [] INT denoms = (25, 10, 5, 1);
  print((ways to make change(denoms, 100), newline))
END

Output:

       +242
Works with: ALGOL 68G version Any - tested with release 2.8.4
Translation of: Haskell
#
   Rosetta Code "Count the coins"
   This uses what I believe are the ideas behind the "much faster, probably
   harder to read" Haskell version.  
#

BEGIN
   PROC ways to make change = ([] INT denoms, INT amount) LONG INT:
   BEGIN
      [0:amount] LONG INT counts, new counts;

      FOR i FROM 0 TO amount DO counts[i] := (i = 0 | 1 | 0) OD;

      FOR i FROM LWB denoms TO UPB denoms DO
         INT denom = denoms[i];
         FOR j FROM 0 TO amount DO new counts[j] := 0 OD;
         FOR j FROM 0 TO amount DO
            IF LONG INT count = counts[j]; count > 0 THEN
               FOR k FROM j + denom BY denom TO amount DO
                  new counts[k] +:= count
               OD
            FI;
            counts[j] +:= new counts[j]
         OD
      OD;
      counts[amount]
   END;

   print((ways to make change((1, 5, 10, 25), 100), newline));
   print((ways to make change((1, 5, 10, 25, 50, 100), 10000), newline));
   print((ways to make change((1, 5, 10, 25, 50, 100), 100000), newline))
END

Output:

                                +242
                       +139946140451
                  +13398445413854501

AppleScript

Translation of: Phix
-- All input values must be integers and multiples of the same monetary unit.
on countCoins(amount, denominations)
    -- Potentially long list of counters, initialised with 1 (result for amount 0) and 'amount' zeros.
    script o
        property counters : {1}
    end script
    repeat amount times
        set end of o's counters to 0
    end repeat
    
    -- Less labour-intensive alternative to the following repeat's c = 1 iteration.
    set coinValue to beginning of denominations
    repeat with n from (coinValue + 1) to (amount + 1) by coinValue
        set item n of o's counters to 1
    end repeat
    
    repeat with c from 2 to (count denominations)
        set coinValue to item c of denominations
        repeat with n from (coinValue + 1) to (amount + 1)
            set item n of o's counters to (item n of o's counters) + (item (n - coinValue) of o's counters)
        end repeat
    end repeat
    
    return end of o's counters
end countCoins

-- Task calls:
set c1 to countCoins(100, {25, 10, 5, 1})
set c2 to countCoins(1000 * 100, {100, 50, 25, 10, 5, 1})
return {c1, c2}
Output:
{242, 13398445413854501}

Applesoft BASIC

Translation of: Commodore BASIC
C=0:M=100:F=25:T=10:S=5:Q=INT(M/F):FORI=0TOQ:D=INT((M-I*F)/T):FORJ=0TOD:N=INT((M-J*T)/S):FORK=0TON:P=M-K*S:FORL=0TOPSTEPS:C=C+(L+K*S+J*T+I*F=M):NEXTL,K,J,I:?C;

Arturo

changes: function [amount coins][
	ways: map 0..amount+1 [x]-> 0
	ways\0: 1

	loop coins 'coin [
		loop coin..amount 'j ->
			set ways j (get ways j) + get ways j-coin
	]

	ways\[amount]
]
 
print changes 100 [1 5 10 25]
print changes 100000 [1 5 10 25 50 100]

AutoHotkey

Translation of: Go
Works with: AutoHotkey_L
countChange(amount){
	return cc(amount, 4)
}

cc(amount, kindsOfCoins){
	if ( amount == 0 )
		return 1
	if ( amount < 0 ) || ( kindsOfCoins == 0 )
		return 0
	return cc(amount, kindsOfCoins-1)
	    +  cc(amount - firstDenomination(kindsOfCoins), kindsOfCoins)
}

firstDenomination(kindsOfCoins){
	return [1, 5, 10, 25][kindsOfCoins]
}
MsgBox % countChange(100)

AWK

Iterative implementation, derived from Run BASIC:

#!/usr/bin/awk -f

BEGIN {
    print cc(100)
    exit
}

function cc(amount, coins,    numPennies, numNickles, numQuarters, p, n, d, q, s, count) {
    numPennies = amount
    numNickles = int(amount / 5)
    numDimes = int(amount / 10)
    numQuarters = int(amount / 25)
    
    count = 0
    for (p = 0; p <= numPennies; p++) {
        for (n = 0; n <= numNickles; n++) {
            for (d = 0; d <= numDimes; d++) {
                for (q = 0; q <= numQuarters; q++) {
                    s = p + n * 5 + d * 10 + q * 25;
                    if (s == 100) count++;
                }
            }
        }
    }
    return count;
}

Run time:

time ./change-itr.awk
242

real	0m0.065s
user	0m0.063s
sys	0m0.002s

Recursive implementation (derived from Scheme example):

#!/usr/bin/awk -f

BEGIN {
    COINSEP = ", "
    coins = 1 COINSEP 5 COINSEP 10 COINSEP 25
    print cc(100, coins)
    exit
}

function cc(amt, coins) {
    if (length(coins) == 0) return 0
    if (amt < 0) return 0
    if (amt == 0) return 1
    return cc(amt, tail(coins)) + cc(amt - head(coins), coins)
}

function tail(coins,    koins, s, c) {
    split(coins, koins, COINSEP)
    s = ""
    for (c = 2; c <= length(koins); c++) s = s (s == "" ? "" : COINSEP) koins[c]
    return s;
}

function head(coins,    koins) {
    split(coins, koins, COINSEP)
    return koins[1]
}

Run time:

time ./change-rec.awk 
242

real	0m0.081s 
user	0m0.079s
sys	0m0.002s

While the recursive version is slower for small amounts, about 2 bucks it gets faster than the iterative version, at least until is segfaults from exhausting the stack.

BBC BASIC

Non-recursive solution:

      DIM uscoins%(3)
      uscoins%() = 1, 5, 10, 25
      PRINT FNchange(100, uscoins%()) " ways of making $1"
      PRINT FNchange(1000, uscoins%()) " ways of making $10"
      
      DIM ukcoins%(7)
      ukcoins%() = 1, 2, 5, 10, 20, 50, 100, 200
      PRINT FNchange(100, ukcoins%()) " ways of making £1"
      PRINT FNchange(1000, ukcoins%()) " ways of making £10"
      END
      
      DEF FNchange(sum%, coins%())
      LOCAL C%, D%, I%, N%, P%, Q%, S%, table()
      C% = 0
      N% = DIM(coins%(),1) + 1
      FOR I% = 0 TO N% - 1
        D% = coins%(I%)
        IF D% <= sum% IF D% >= C% C% = D% + 1
      NEXT
      C% *= N%
      DIM table(C%-1)
      FOR I% = 0 TO N%-1 : table(I%) = 1 : NEXT
      
      P% = N%
      FOR S% = 1 TO sum%
        FOR I% = 0 TO N% - 1
          IF I% = 0 IF P% >= C% P% = 0
          IF coins%(I%) <= S% THEN
            Q% = P% - coins%(I%) * N%
            IF Q% >= 0 table(P%) = table(Q%) ELSE table(P%) = table(Q% + C%)
          ENDIF
          IF I% table(P%) += table(P% - 1)
          P% += 1
        NEXT
      NEXT
      = table(P%-1)

Output (BBC BASIC does not have large enough integers for the optional task):

       242 ways of making $1
    142511 ways of making $10
      4563 ways of making £1
 321335886 ways of making £10

C

Using some crude 128-bit integer type.

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>

// ad hoc 128 bit integer type; faster than using GMP because of low
// overhead
typedef struct { uint64_t x[2]; } i128;

// display in decimal
void show(i128 v) {
	uint32_t x[4] = {v.x[0], v.x[0] >> 32, v.x[1], v.x[1] >> 32};
	int i, j = 0, len = 4;
	char buf[100];
	do {
		uint64_t c = 0;
		for (i = len; i--; ) {
			c = (c << 32) + x[i];
			x[i] = c / 10, c %= 10;
		}

		buf[j++] = c + '0';
		for (len = 4; !x[len - 1]; len--);
	} while (len);

	while (j--) putchar(buf[j]);
	putchar('\n');
}

i128 count(int sum, int *coins)
{
	int n, i, k;
	for (n = 0; coins[n]; n++);

	i128 **v = malloc(sizeof(int*) * n);
	int *idx = malloc(sizeof(int) * n);

	for (i = 0; i < n; i++) {
		idx[i] = coins[i];
		// each v[i] is a cyclic buffer
		v[i] = calloc(sizeof(i128), coins[i]);
	}

	v[0][coins[0] - 1] = (i128) {{1, 0}};

	for (k = 0; k <= sum; k++) {
		for (i = 0; i < n; i++)
			if (!idx[i]--) idx[i] = coins[i] - 1;

		i128 c = v[0][ idx[0] ];

		for (i = 1; i < n; i++) {
			i128 *p = v[i] + idx[i];

			// 128 bit addition
			p->x[0] += c.x[0];
			p->x[1] += c.x[1];
			if (p->x[0] < c.x[0]) // carry
				p->x[1] ++;
			c = *p;
		}
	}

	i128 r = v[n - 1][idx[n-1]];

	for (i = 0; i < n; i++) free(v[i]);
	free(v);
	free(idx);

	return r;
}

// simple recursive method; slow
int count2(int sum, int *coins)
{
	if (!*coins || sum < 0) return 0;
	if (!sum) return 1;
	return count2(sum - *coins, coins) + count2(sum, coins + 1);
}

int main(void)
{
	int us_coins[] = { 100, 50, 25, 10, 5, 1, 0 };
	int eu_coins[] = { 200, 100, 50, 20, 10, 5, 2, 1, 0 };

	show(count(   100, us_coins + 2));
	show(count(  1000, us_coins));

	show(count(  1000 * 100, us_coins));
	show(count( 10000 * 100, us_coins));
	show(count(100000 * 100, us_coins));

	putchar('\n');

	show(count(     1 * 100, eu_coins));
	show(count(  1000 * 100, eu_coins));
	show(count( 10000 * 100, eu_coins));
	show(count(100000 * 100, eu_coins));

	return 0;
}

output (only the first two lines are required by task):

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4563
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

C#

    // Adapted from http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/
    class Program
    {
        static long Count(int[] C, int m, int n)
        {
            var table = new long[n + 1];
            table[0] = 1;
            for (int i = 0; i < m; i++)
                for (int j = C[i]; j <= n; j++)
                    table[j] += table[j - C[i]];
            return table[n];
        }
        static void Main(string[] args)
        {
            var C = new int[] { 1, 5, 10, 25 };
            int m = C.Length;
            int n = 100;
            Console.WriteLine(Count(C, m, n));  //242
            Console.ReadLine();
        }
    }

C++

#include <iostream>
#include <stack>
#include <vector>

struct DataFrame {
  int sum;
  std::vector<int> coins;
  std::vector<int> avail_coins;
};

int main() {
  std::stack<DataFrame> s;
  s.push({ 100, {}, { 25, 10, 5, 1 } });
  int ways = 0;
  while (!s.empty()) {
    DataFrame top = s.top();
    s.pop();
    if (top.sum < 0) continue;
    if (top.sum == 0) {
      ++ways;
      continue;
    }
    if (top.avail_coins.empty()) continue;
    DataFrame d = top;
    d.sum -= top.avail_coins[0];
    d.coins.push_back(top.avail_coins[0]);
    s.push(d);
    d = top;
    d.avail_coins.erase(std::begin(d.avail_coins));
    s.push(d);
  }
  std::cout << ways << std::endl;
  return 0;
}
Output:
242

Clojure

(def denomination-kind [1 5 10 25])

(defn- cc [amount denominations]
  (cond (= amount 0) 1
        (or (< amount 0) (empty? denominations)) 0
        :else (+ (cc amount (rest denominations))
                 (cc (- amount (first denominations)) denominations))))

(defn count-change
  "Calculates the number of times you can give change with the given denominations."
  [amount denominations]
  (cc amount denominations))

(count-change 15 denomination-kind) ; = 6

COBOL

Translation of: C#
       identification division.
       program-id. CountCoins.

       data division.
       working-storage section.
       77  i                      pic 9(3).
       77  j                      pic 9(3).  
       77  m                      pic 9(3) value 4.
       77  n                      pic 9(3) value 100.  
       77  edited-value           pic z(18).
       01  coins-table            value "01051025".
           05 coin                pic 9(2) occurs 4.
       01  ways-table.
           05 way                 pic 9(18) occurs 100.

       procedure division.
       main.
           perform calc-count
           move way(n) to edited-value
           display function trim(edited-value)
           stop run
           .
       calc-count.
           initialize ways-table
           move 1 to way(1)
           perform varying i from 1 by 1 until i > m
              perform varying j from coin(i) by 1 until j > n
                 add way(j - coin(i)) to way(j)
              end-perform
           end-perform
           .
Output:
242

Coco

Translation of: Python
changes = (amount, coins) ->
    ways = [1].concat [0] * amount
    for coin of coins
        for j from coin to amount
            ways[j] += ways[j - coin]
    ways[amount]
 
console.log changes 100, [1 5 10 25]

Commodore BASIC

Example 1: Base example in Commodore BASIC (works on PET, C64, VIC20, etc.)

This example is based on the Spectrum ZX BASIC example found below. Direct copy of that algorithm and executed on an emulated Commodore 64 in VICE resulted in a timed performance of 46 minutes and 37 seconds (46:37) as measured by the C64 BASIC system clock (TIME$ or TI$, times are approximate within a few seconds). Some improvements were made as follows:

  1. Reversed the order of the loops to start counting with the largest denomination > smallest denomination. Result: 44:45
  2. It makes no sense to check with anything other than a multiple of 5 pennies, since the other denominations value a multiple of 5. Adding "step 5" to the penny for loop skips over a good portion of useless iteration. Result: about 9:44.
  3. Not printing any of the individual results speeds up total time to 9:30.
  4. Removing the specific variables used in the NEXT statements helps the interpreter speed up. Result: 9:10.
  5. Now that the denominations were reordered, it makes sense that each sub-loop with the next lower denomination should loop only through the remaining money not accounted for by the larger denomination. Result: 2:12.


5 m=100:rem money = $1.00 or 100 pennies.
10 print chr$(147);chr$(14);"This program will calculate the number"
11 print "of combinations of 'change' that can be"
12 print "given for a $1 bill."
13 print:print "The coin values are:"
14 print "0.01 = Penny":print "0.05 = Nickle"
15 print "0.10 = Dime":print "0.25 = Quarter"
16 print
20 print "Would you like to see each combination?"
25 get k$:yn=(k$="y"):if k$="" then 25
100 p=m:ti$="000000"
130 q=int(m/25)
140 count=0:ps=1
147 if yn then print "Count  P    N    D    Q"
150 for qc=0 to q:d=int((m-qc*25)/10)
160 for dc=0 to d:n=int((m-dc*10)/5)
170 for nc=0 to n:p=m-nc*5
180 for pc=0 to p step 5
190 s=pc+nc*5+dc*10+qc*25
200 if s=m then count=count+1:if yn then gosub 1000
210 next:next:next:next
245 en$=ti$
250 print:print count;"different combinations found in"
260 print tab(len(str$(count))+1);
265 print left$(en$,2);":";mid$(en$,3,2);":";right$(en$,2);"."
270 end
1000 print count;tab(6);pc;tab(11);nc;tab(16);dc;tab(21);qc:return

Example 2: Commodore 64 with Screen Blanking

Make the following changes on a Commodore 64 to enable screen blanking. This will give the CPU a few extra cycles normally held by the VIC-II. Add line 145 and change line 245 as shown.

Enabling screen blanking (and therefore not printing each result) results in a total time of 1:44.

145 if not yn then poke 53265,peek(53265) and 239
245 en$=ti$:if not yn then poke 53265,peek(53265) or 16

Example 3: Commodore 128 with VIC-II blanking, 2MHz fast mode.

Similar to above, however the Commodore 128 is capable of using a faster clock speed at the expense of any VIC-II graphics display. Timed result is 1:18. Add/change the following lines on the Commodore 128:

145 if not yn then fast
245 en$=ti$:if not yn then slow

Common Lisp

Recursive Version With Cache

(defun count-change (amount coins
                    &optional
                    (length (1- (length coins)))
                    (cache  (make-array (list (1+ amount) (length coins))
                                        :initial-element nil)))
  (cond ((< length 0) 0)
        ((< amount 0) 0)
        ((= amount 0) 1)
        (t (or (aref cache amount length)
               (setf (aref cache amount length)
                     (+ (count-change (- amount (first coins)) coins length cache)
                        (count-change amount (rest coins) (1- length) cache)))))))

; (compile 'count-change) ; for CLISP

(print (count-change 100 '(25 10 5 1)))		   ; = 242
(print (count-change 100000 '(100 50 25 10 5 1)))  ; = 13398445413854501
(terpri)

Iterative Version

(defun count-change (amount coins &aux (ways (make-array (1+ amount) :initial-element 0)))
  (setf (aref ways 0) 1)
  (loop for coin in coins do
        (loop for j from coin upto amount
              do (incf (aref ways j) (aref ways (- j coin)))))
  (aref ways amount))

D

Basic Version

Translation of: Go
import std.stdio, std.bigint;

auto changes(int amount, int[] coins) {
    auto ways = new BigInt[amount + 1];
    ways[0] = 1;
    foreach (coin; coins)
        foreach (j; coin .. amount + 1)
            ways[j] += ways[j - coin];
    return ways[$ - 1];
}

void main() {
    changes(   1_00, [25, 10, 5, 1]).writeln;
    changes(1000_00, [100, 50, 25, 10, 5, 1]).writeln;
}
Output:
242
13398445413854501

Safe Ulong Version

This version is very similar to the precedent, but it uses a faster ulong type, and performs a checked sum to detect overflows at run-time.

import std.stdio, core.checkedint;

auto changes(int amount, int[] coins, ref bool overflow) {
    auto ways = new ulong[amount + 1];
    ways[0] = 1;
    foreach (coin; coins)
        foreach (j; coin .. amount + 1)
            ways[j] = ways[j].addu(ways[j - coin], overflow);
    return ways[amount];
}

void main() {
    bool overflow = false;
    changes(    1_00, [25, 10, 5, 1], overflow).writeln;
    if (overflow)
        "Overflow".puts;
    overflow = false;
    changes( 1000_00, [100, 50, 25, 10, 5, 1], overflow).writeln;
    if (overflow)
        "Overflow".puts;
}

The output is the same.

Faster Version

Translation of: C
import std.stdio, std.bigint;

BigInt countChanges(in int amount, in int[] coins) pure /*nothrow*/ {
    immutable n = coins.length;
    int cycle;
    foreach (immutable c; coins)
        if (c <= amount && c >= cycle)
            cycle = c + 1;
    cycle *= n;
    auto table = new BigInt[cycle];
    table[0 .. n] = 1.BigInt;

    int pos = n;
    foreach (immutable s; 1 .. amount + 1) {
        foreach (immutable i; 0 .. n) {
            if (i == 0 && pos >= cycle)
                pos = 0;
            if (coins[i] <= s) {
                immutable int q = pos - (coins[i] * n);
                table[pos] = (q >= 0) ? table[q] : table[q + cycle];
            }
            if (i)
                table[pos] += table[pos - 1];
            pos++;
        }
    }

    return table[pos - 1];
}

void main() {
    immutable usCoins = [100, 50, 25, 10, 5, 1];
    immutable euCoins = [200, 100, 50, 20, 10, 5, 2, 1];

    foreach (immutable coins; [usCoins, euCoins]) {
        countChanges(     1_00, coins[2 .. $]).writeln;
        countChanges(  1000_00, coins).writeln;
        countChanges( 10000_00, coins).writeln;
        countChanges(100000_00, coins).writeln;
        writeln;
    }
}
Output:
242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

128-bit Version

A much faster version that mixes high-level and low-level style programming. This version uses basic 128-bit unsigned integers, like the C version. The output is the same as the second D version.

Translation of: C
import std.stdio, std.bigint, std.algorithm, std.conv, std.functional;

struct Ucent { /// Simplified 128-bit integer (like ucent).
    ulong hi, lo;
    static immutable one = Ucent(0, 1);

    void opOpAssign(string op="+")(in ref Ucent y) pure nothrow @nogc @safe {
        this.hi += y.hi;
        if (this.lo >= ~y.lo)
            this.hi++;
        this.lo += y.lo;
    }

    string toString() const /*pure nothrow @safe*/ {
        return text((this.hi.BigInt << 64) + this.lo);
    }
}

Ucent countChanges(in int amount, in int[] coins) pure nothrow {
    immutable n = coins.length;

    // Points to a cyclic buffer of length coins[i]
    auto p = new Ucent*[n];
    auto q = new Ucent*[n]; // iterates it.
    auto buf = new Ucent[coins.sum];

    p[0] = buf.ptr;
    foreach (immutable i; 0 .. n) {
        if (i)
            p[i] = coins[i - 1] + p[i - 1];
        *p[i] = Ucent.one;
        q[i] = p[i];
    }

    Ucent prev;
    foreach (immutable j; 1 .. amount + 1)
        foreach (immutable i; 0 .. n) {
            q[i]--;
            if (q[i] < p[i])
                q[i] = p[i] + coins[i] - 1;
            if (i)
                *q[i] += prev;
            prev = *q[i];
        }

    return prev;
}

void main() {
    immutable usCoins = [100, 50, 25, 10, 5, 1];
    immutable euCoins = [200, 100, 50, 20, 10, 5, 2, 1];

    foreach (immutable coins; [usCoins, euCoins]) {
        countChanges(     1_00, coins[2 .. $]).writeln;
        countChanges(  1000_00, coins).writeln;
        countChanges( 10000_00, coins).writeln;
        countChanges(100000_00, coins).writeln;
        writeln;
    }
}

Printing Version

This version prints all the solutions (so it can be used on the smaller input):

import std.stdio, std.conv, std.string, std.algorithm, std.range;

void printChange(in uint tot, in uint[] coins)
in {
    assert(coins.isSorted);
} body {
    auto freqs = new uint[coins.length];

    void inner(in uint curTot, in size_t start) {
        if (curTot == tot)
            return writefln("%-(%s %)",
                            zip(coins, freqs)
                            .filter!(cf => cf[1] != 0)
                            .map!(cf => format("%u:%u", cf[])));

        foreach (immutable i; start .. coins.length) {
            immutable ci = coins[i];
            for (auto v = (freqs[i] + 1) * ci; v <= tot; v += ci)
                if (curTot + v <= tot) {
                    freqs[i] += v / ci;
                    inner(curTot + v, i + 1);
                    freqs[i] -= v / ci;
                }
        }
    }

    inner(0, 0);
}

void main() {
    printChange(1_00, [1, 5, 10, 25]);
}
Output:
1:5 5:1 10:4 25:2
1:5 5:1 10:9
1:5 5:2 10:1 25:3
1:5 5:2 10:6 25:1
1:5 5:3 10:3 25:2
1:5 5:3 10:8
1:5 5:4 10:5 25:1
1:5 5:4 25:3
1:5 5:5 10:2 25:2
1:5 5:5 10:7
1:5 5:6 10:4 25:1
1:5 5:7 10:1 25:2
...
5:11 10:2 25:1
5:12 10:4
5:13 10:1 25:1
5:14 10:3
5:15 25:1
5:16 10:2
5:18 10:1
5:20
10:5 25:2
10:10
25:4

Dart

Simple recursive version plus cached version using a map.

Dart 1 version:

var cache = new Map();

main() {
    var stopwatch = new Stopwatch()..start();

    // use the brute-force recursion for the small problem
    int amount = 100;
    list coinTypes = [25,10,5,1];
    print (coins(amount,coinTypes).toString() + " ways for $amount using $coinTypes coins.");

    // use the cache version for the big problem
    amount = 100000;
    coinTypes = [100,50,25,10,5,1];
    print (cachedCoins(amount,coinTypes).toString() + " ways for $amount using $coinTypes coins.");

    stopwatch.stop();
    print ("... completed in " + (stopwatch.elapsedMilliseconds/1000).toString() + " seconds");
}


coins(int amount, list coinTypes) {
    int count = 0;

    if(coinTypes.length == 1) return (1);   // just pennies available, so only one way to make change

    for(int i=0; i<=(amount/coinTypes[0]).toInt(); i++){                // brute force recursion
      count += coins(amount-(i*coinTypes[0]),coinTypes.sublist(1));     // sublist(1) is like lisp's '(rest ...)'
    }

    // uncomment if you want to see intermediate steps
    //print("there are " + count.toString() +" ways to count change for ${amount.toString()} using ${coinTypes} coins.");
    return(count);
  }


  cachedCoins(int amount, list coinTypes) {
      int count = 0;

      // this is more efficient, looks at last two coins.  but not fast enough for the optional exercise.
      if(coinTypes.length == 2) return ((amount/coinTypes[0]).toInt() + 1);

      var key = "$amount.$coinTypes";         // lookes like "100.[25,10,5,1]"
      var cacheValue = cache[key];            // check whether we have seen this before

      if(cacheValue != null) return(cacheValue);

      count = 0;
      // same recursion as simple method, but caches all subqueries too
      for(int i=0; i<=(amount/coinTypes[0]).toInt(); i++){
        count += cachedCoins(amount-(i*coinTypes[0]),coinTypes.sublist(1));     // sublist(1) is like lisp's '(rest ...)'
      }

      cache[key] = count;                     // add this to the cache
      return(count);
    }
Output:
242 ways for 100 using [25, 10, 5, 1] coins.
13398445413854501 ways for 100000 using [100, 50, 25, 10, 5, 1] coins.
... completed in 3.604 seconds

Dart 2 version:

/// Provides the same result and performance as the Dart 1 version
/// but using the Dart 2 specifications.
Map<String, int> cache = {};

void main() {
  Stopwatch stopwatch = Stopwatch()..start();

  /// Use the brute-force recursion for the small problem
  int amount = 100;
  List<int> coinTypes = [25,10,5,1];
  print ("${coins(amount,coinTypes)} ways for $amount using $coinTypes coins.");

  /// Use the cache version for the big problem
  amount = 100000;
  coinTypes = [100,50,25,10,5,1];
  print ("${cachedCoins(amount,coinTypes)} ways for $amount using $coinTypes coins.");

  stopwatch.stop();
  print ("... completed in ${stopwatch.elapsedMilliseconds/1000} seconds");

}

int cachedCoins(int amount, List<int> coinTypes) {
  int count = 0;

  /// This is more efficient, looks at last two coins.
  /// But not fast enough for the optional exercise.
  if(coinTypes.length == 2) return (amount ~/ coinTypes[0] + 1);

  /// Looks like "100.[25,10,5,1]"
  String key = "$amount.$coinTypes";
  /// Check whether we have seen this before
  var cacheValue = cache[key];

  if(cacheValue != null) return(cacheValue);

  count = 0;
  /// Same recursion as simple method, but caches all subqueries too
  for(int i=0; i<=amount ~/ coinTypes[0]; i++){
    count += cachedCoins(amount-(i*coinTypes[0]),coinTypes.sublist(1));     // sublist(1) is like lisp's '(rest ...)'
  }

  /// add this to the cache
  cache[key] = count;
  return count;
}

int coins(int amount, List<int> coinTypes) {
  int count = 0;

  /// Just pennies available, so only one way to make change
  if(coinTypes.length == 1) return (1);

  /// Brute force recursion
  for(int i=0; i<=amount ~/ coinTypes[0]; i++){
    /// sublist(1) is like lisp's '(rest ...)'
    count += coins(amount - (i*coinTypes[0]),coinTypes.sublist(1));
  }

  /// Uncomment if you want to see intermediate steps
  /// print("there are " + count.toString() +" ways to count change for ${amount.toString()} using ${coinTypes} coins.");
  return count;
}
Output:
242 ways for 100 using [25, 10, 5, 1] coins.
13398445413854501 ways for 100000 using [100, 50, 25, 10, 5, 1] coins.
... completed in 2.921 seconds

Process finished with exit code 0

Delphi

Translation of: C#
program Count_the_coins;

{$APPTYPE CONSOLE}

function Count(c: array of Integer; m, n: Integer): Integer;
var
  table: array of Integer;
  i, j: Integer;
begin
  SetLength(table, n + 1);
  table[0] := 1;
  for i := 0 to m - 1 do
    for j := c[i] to n do
      table[j] := table[j] + table[j - c[i]];
  Exit(table[n]);
end;

var
  c: array of Integer;
  m, n: Integer;

begin
  c := [1, 5, 10, 25];

  m := Length(c);
  n := 100;
  Writeln(Count(c, m, n));  //242
  Readln;
end.
Output:
242

Draco

proc main() void:
    [4]byte coins = (1, 5, 10, 25);
    [101]byte tab;
    word m, n;

    for n from 1 upto 100 do tab[n] := 0 od;
    tab[0] := 1;

    for m from 0 upto 3 do
        for n from coins[m] upto 100 do
            tab[n] := tab[n] + tab[n - coins[m]]
        od
    od;

    writeln(tab[100])
corp
Output:
242

Dyalect

func countCoins(coins, n) {
    var xs = Array.Empty(n + 1, 0)
    xs[0] = 1
    for c in coins {
        var cj = c
        while cj <= n {
            xs[cj] += xs[cj - c]
            cj += 1
        }
    }
    return xs[n]
}

var coins = [1, 5, 10, 25]
print(countCoins(coins, 100))
Output:
242

EasyLang

len cache[] 100000 * 7 + 6
val[] = [ 1 5 10 25 50 100 ]
func count sum kind .
   if sum = 0
      return 1
   .
   if sum < 0 or kind = 0
      return 0
   .
   chind = sum * 7 + kind
   if cache[chind] > 0
      return cache[chind]
   .
   r2 = count (sum - val[kind]) kind
   r1 = count sum (kind - 1)
   r = r1 + r2
   cache[chind] = r
   return r
.
print count 100 4
print count 10000 6
print count 100000 6
# this is not exact, since numbers
# are doubles and r > 2^53

EchoLisp

Recursive solution using memoization, adapted from CommonLisp and Racket.

(lib 'compile) ;; for (compile)
(lib 'bigint)  ;; integer results > 32 bits
(lib 'hash)    ;; hash table

;; h-table
(define Hcoins (make-hash))

;; the function to memoize
(define (sumways cents coins)
	(+ (ways cents (cdr coins)) (ways (- cents (car coins)) coins)))
	
;; accelerator : ways (cents, coins) = ways ((cents  - cents % 5) , coins)
(define (ways cents coins)
  (cond ((null? coins) 0)
        ((negative? cents) 0)
        ((zero? cents) 1)
        ((eq? coins c-1) 1) ;; if coins = (1) --> 1
        (else (hash-ref! Hcoins (list (- cents (modulo cents 5)) coins) sumways))))

(compile 'ways) ;; speed-up things
Output:
(define change '(25 10 5 1))
(define c-1 (list-tail change -1)) ;; pointer to (1)
(ways 100 change)
     242

(define change '(100 50 25 10 5 1))
(define c-1 (list-tail change -1))
(for ((i (in-range 0 200001 20000))) 
    (writeln i (time (ways i change)) (hash-count Hcoins)))


;; iterate cents = 20000, 40000, ..
;; cents ((time (msec) number-of-ways) number-of-entries-in-h-table

20000      (350 4371565890901)         9398    
40000      (245 138204514221801)       18798    
60000      (230 1045248220992701)      28198    
80000      (255 4395748062203601)      37598    
100000     (234 13398445413854501)     46998    
120000     (230 33312577651945401)     56398    
140000     (292 71959878152476301)     65798    
160000     (736 140236576291447201)     75198    
180000     (237 252625397444858101)     84598    
200000     (240 427707562988709001)     93998    

;; One can see that the time is linear, and the h-table size reasonably small

change 
     (100 50 25 10 5 1)
(ways 100000 change)
     13398445413854501

EDSAC order code

The program solves the first task for the US dollar and UK pound, using an algorithm copied from the C# and Delphi solutions. The second task is not attempted.

Note: When the table is initialized, not only must the first entry be set to 1, but the other entries must be set to 0. It seems that the C# and Delphi solutions rely on the compiler to do this. In other languages, it may need to be done by the program.

["Count the coins" problem for Rosetta Code.]
[EDSAC program, Initial Orders 2.]

            T51K  P56F [G parameter: print subroutine]
            T54K  P94F [C parameter: coins subroutine]
            T47K P200F [M parameter: main routine]

[========================== M parameter ===============================]
            E25K TM GK 
[Parameter block for US coins. For convenience, all numbers
 are in the address field, e.g. 25 cents is P25F not P12D.]
      [0]   UF SF   [2-letter ID]
            P100F   [amount to be made with coins]
            P4F     [number of coin values]
            P1F P5F P10F P25F [list of coin values]
      [8]   P@      [address of US parameter block]
[Parameter block for UK coins]
      [9]   UF KF
            P100F
            P7F
            P1F P2F P5F P10F P20F P50F P100F
     [20]   P9@     [address of UK parameter block]
[Enter with acc = 0]
     [21]   A8@     [load address of parameter block for US coins]
            T4F     [pass to subroutine in 4F]
     [23]   A23@    [call subroutine to calculate and print result]
            G13C
            A20@    [same for UK coins]
            T4F
     [27]   A27@
            G13C
            ZF      [halt program]

[========================== C parameter ===============================]
[Subroutine to calculate and print the result for the given amount and
 set of coins. Address of parameter block (see above) is passed in 4F.]
 
            E25K TC GK
      [0]   SF      [S order for start of coin list]
      [1]   A1023F  [start table at top of memory and work downwarda]
      [2]   PF      [S order for exclusive end of coin list]
      [3]   P2F     [to increment address by 2]
      [4]   OF      [(1) add to address to make O order
                     (2) add to A order to make T order with same address]
      [5]   SF      [add to address to make S order]
      [6]   K4095F  [add to S order to make A order, dec address]
      [7]   K2048F  [set teleprinter to letters]
      [8]   #F      [set teleprinter to figures]
      [9]   !F      [space character]
     [10]   @F      [carriage return]
     [11]   &F      [line feed]
     [12]   K4096F  [teleprinter null]
[Subroutine entry. In this EDSAC program, the table used
 in the algorithm grows downward from the top of memory.]
     [13]   A3F     [plant jump back to caller, as usual]
            T89@
            A4F     [load address of parameter block]
            A3@     [skip 2-letter ID]
            A5@     [make S order for amount]
            U27@    [plant in code]
            A3@     [make S order for first coin value]
            U@      [store it]
            A6@     [make A order for number of coins]
            T38@    [plant in code]
            A2F     [load 1 (in address field)]
     [24]   T1023F  [store at start of table]
[Set all other table entries to 0]
            A24@
            T32@
     [27]   SF      [acc := -amount]
     [28]   TF      [set negative count in 0F]
            A32@    [decrement address in manufactured order]
            S2F
            T32@
     [32]   TF      [manufactured: set table entry to 0]
            AF      [update negative count]
            A2F
            G28@    [loop until count = 0]
[Here acc = 0. Manufactured order (4 lines up) is T order
 for inclusive end of table; this is used again below.]
            A@      [load S order for first coin value]
            U43@    [plant in code]
     [38]   AF      [make S order for exclusive end of coin list]
            T2@     [store for comparison]
[Start of outer loop, round coin values]
     [40]   TF      [clear acc]
            A1@     [load A order for start of table]
            U48@    [plant in code]
     [43]   SF      [manufactured order: subtract coin value]
[Start of inner loop, round table entries]
     [44]   U47@    [plant A order in code]
            A4@     [make T order for same address]
            T49@    [plant in code]
[The next 3 orders are manufactured at run time]
     [47]   AF      [load table entry]
     [48]   AF      [add earlier table entry]
     [49]   TF      [update table entry]
            A32@    [load T order for inclusive end of table]
            S49@    [reached end of table?]
            E60@    [if yes, jump out of inner loop]
            TF      [clear acc]
            A48@    [update the 3 manufactured instructions]
            S2F
            T48@
            A47@
            S2F
            G44@    [always loops back, since A < 0]
[End of inner loop]
     [60]   TF      [clear acc]
            A43@    [update S order for coin value]
            A2F
            U43@
            S2@     [reached exclusive end?]
            G40@    [if no, loop back]
[End of outer loop]
[Here with acc = 0 and result at end of table]
[Value is in address field, so shift 1 right for printing]
            A32@    [load T order for end of tab;e]
            S4@     [make A order for same address]
            T79@    [plant in code]
            A4F     [load address of parameter block]
            A4@     [make O order for 1st char of ID]
            U75@    [plant in code]
            A2F     [same for 2nd char]
            T76@
            O7@     [set teleprinter to letters]
     [75]   OF      [print ID, followed by space]
     [76]   OF O9@
            O8@     [set teleprinter to figures]
     [79]   AF      [maunfactured order to load result]
            RD      [shift 1 right for printing]
            TF      [pass to print routine]
            A9@     [replace leading 0's with space]
            T1F
     [84]   A84@    [call print routine]
            GG
            O10@ O11@  [print CR, LF]
            O12@    [print null to flush teleprinter buffer]
     [89]   ZF      [replaced by jump back to caller]

[============================= G parameter ===============================]
            E25K TG GK 
[Subroutine to print non-negative 17-bit integer. Always prints 5 chars.
 Caller specifies character for leading 0 (typically 0, space or null).
 Parameters: 0F = integer to be printed (not preserved)
             1F = character for leading zero (preserved)
 Workspace: 4F..7F, 38 locations]
            A3FT34@A1FT7FS35@T6FT4#FAFT4FH36@V4FRDA4#FR1024FH37@E23@O7FA2F
            T6FT5FV4#FYFL8FT4#FA5FL1024FUFA6FG16@OFTFT7FA6FG17@ZFP4FZ219DTF

[========================== M parameter again ===============================]
            E25K TM GK
            E21Z    [define entry point]
            PF      [enter with acc = 0]
Output:
US   242
UK  4563


Elixir

Recursive Dynamic Programming solution in Elixir

defmodule Coins do
  def find(coins,lim) do
    vals = Map.new(0..lim,&{&1,0}) |> Map.put(0,1)
    count(coins,lim,vals)
      |> Map.values
      |> Enum.max
      |> IO.inspect
  end
  
  defp count([],_,vals), do: vals
  defp count([coin|coins],lim,vals) do
    count(coins,lim,ways(coin,coin,lim,vals))
  end
  
  defp ways(num,_coin,lim,vals) when num > lim, do: vals
  defp ways(num, coin,lim,vals) do
    ways(num+1,coin,lim,ad(coin,num,vals))
  end
  
  defp ad(a,b,c), do: Map.put(c,b,c[b]+c[b-a])
end

Coins.find([1,5,10,25],100)
Coins.find([1,5,10,25,50,100],100_000)
Output:
242
13398445413854501

Erlang

-module(coins).
-compile(export_all).

count(Amount, Coins) ->
    {N,_C} = count(Amount, Coins, dict:new()),
    N.

count(0,_,Cache) ->
    {1,Cache};
count(N,_,Cache) when N < 0 ->
    {0,Cache};
count(_N,[],Cache) ->
    {0,Cache};
count(N,[C|Cs]=Coins,Cache) ->
    case dict:is_key({N,length(Coins)},Cache) of
        true -> 
            {dict:fetch({N,length(Coins)},Cache), Cache};
        false ->
            {N1,C1} = count(N-C,Coins,Cache),
            {N2,C2} = count(N,Cs,C1),
            {N1+N2,dict:store({N,length(Coins)},N1+N2,C2)}
    end.

print(Amount, Coins) ->
    io:format("~b ways to make change for ~b cents with ~p coins~n",[count(Amount,Coins),Amount,Coins]).

test() ->
    A1 = 100, C1 = [25,10,5,1],
    print(A1,C1),
    A2 = 100000, C2 = [100, 50, 25, 10, 5, 1],
    print(A2,C2).
Output:
42> coins:test().
242 ways to make change for 100 cents with [25,10,5,1] coins
13398445413854501 ways to make change for 100000 cents with [100,50,25,10,5,1] coins
ok

F#

Translation of: OCaml

Forward iteration, which can also be seen in Scala.

let changes amount coins =
    let ways = Array.zeroCreate (amount + 1)
    ways.[0] <- 1L
    List.iter (fun coin ->
        for j = coin to amount do ways.[j] <- ways.[j] + ways.[j - coin]
    ) coins
    ways.[amount]
 
[<EntryPoint>]
let main argv = 
    printfn "%d" (changes    100 [25; 10; 5; 1]);
    printfn "%d" (changes 100000 [100; 50; 25; 10; 5; 1]);
    0
Output:
242
13398445413854501

Factor

USING: combinators kernel locals math math.ranges sequences sets sorting ;
IN: rosetta.coins

<PRIVATE
! recursive-count uses memoization and local variables.
! coins must be a sequence.
MEMO:: recursive-count ( cents coins -- ways )
    coins length :> types
    {
        ! End condition: 1 way to make 0 cents.
        { [ cents zero? ] [ 1 ] }
        ! End condition: 0 ways to make money without any coins.
        { [ types zero? ] [ 0 ] }
        ! Optimization: At most 1 way to use 1 type of coin.
        { [ types 1 number= ] [
            cents coins first mod zero? [ 1 ] [ 0 ] if
        ] }
        ! Find all ways to use the first type of coin.
        [
            ! f = first type, r = other types of coins.
            coins unclip-slice :> f :> r
            ! Loop for 0, f, 2*f, 3*f, ..., cents.
            0 cents f <range> [
                ! Recursively count how many ways to make remaining cents
                ! with other types of coins.
                cents swap - r recursive-count
            ] [ + ] map-reduce          ! Sum the counts.
        ]
    } cond ;
PRIVATE>

! How many ways can we make the given amount of cents
! with the given set of coins?
: make-change ( cents coins -- ways )
    members [ ] inv-sort-with   ! Sort coins in descending order.
    recursive-count ;

From the listener:

USE: rosetta.coins
( scratchpad ) 100 { 25 10 5 1 } make-change .
242
( scratchpad ) 100000 { 100 50 25 10 5 1 } make-change .
13398445413854501

This algorithm is slow. A test machine needed 1 minute to run 100000 { 100 50 25 10 5 1 } make-change . and get 13398445413854501. The same machine needed less than 1 second to run the Common Lisp (SBCL), Ruby (MRI) or Tcl (tclsh) programs and get the same answer.

One might make use of the rosetta-code.count-the-coins vocabulary as shown:

IN: scratchpad [ 100000 { 1 5 10 25 50 100 } make-change . ] time
13398445413854501
Running time: 0.020869274 seconds

For reference, the implementation is shown next.

USING: arrays locals math math.ranges sequences sets sorting ;
IN: rosetta-code.count-the-coins

<PRIVATE

:: (make-change) ( cents coins -- ways )
    cents 1 + 0 <array> :> ways
    1 ways set-first
    coins [| coin |
        coin cents [a,b] [| j |
            j coin - ways nth j ways [ + ] change-nth
        ] each
    ] each ways last ;

PRIVATE>

! How many ways can we make the given amount of cents
! with the given set of coins?
: make-change ( cents coins -- ways )
    members [ ] inv-sort-with (make-change) ;

Or one could implement the algorithm like described in http://www.cdn.geeksforgeeks.org/dynamic-programming-set-7-coin-change.

USE: math.ranges 

:: exchange-count ( seq val -- cnt )
  val 1 + 0 <array> :> tab
  0 :> old!
  1 0 tab set-nth
  seq length iota [
    seq nth old!
    old val [a,b] [| j |
      j old - tab nth
      j tab nth + 
      j tab set-nth
    ] each
  ] each
  val tab nth
;

[ { 1 5 10 25 50 100 } 100000 exchange-count . ] time
13398445413854501
Running time: 0.029163549 seconds

FOCAL

01.10 S C(1)=1;S C(2)=5;S C(3)=10;S C(4)=25
01.20 F N=1,100;S T(N)=0
01.30 S T(0)=1
01.40 F M=1,4;F N=C(M),100;S T(N)=T(N)+T(N-C(M))
01.50 T %3,T(100),!
01.60 Q
Output:
= 242

Forth

\ counting change (SICP section 1.2.2)

: table create does> swap cells + @ ;
table coin-value 0 , 1 , 5 , 10 , 25 , 50 ,

: count-change ( total coin -- n )
  over 0= if
    2drop 1
  else over 0< over 0= or if
    2drop 0
  else
    2dup coin-value - over recurse
    >r 1- recurse r> +
  then then ;

100 5 count-change .

FreeBASIC

Translation from "Dynamic Programming Solution: Python version" on this webside [1]

' version 09-10-2016
' compile with: fbc -s console


Function count(S() As UInteger, n As UInteger) As ULongInt

  Dim As Integer i, j
  ' calculate m from array S()
  Dim As UInteger m = UBound(S) - LBound(S) +1
  Dim As ULongInt x, y

  '' We need n+1 rows as the table is consturcted in bottom up manner using
  '' the base case 0 value case (n = 0)
  Dim As ULongInt table(n +1, m)

  '' Fill the enteries for 0 value case (n = 0)
  For i = 0 To m -1
    table(0, i) = 1
  Next

  '' Fill rest of the table enteries in bottom up manner
  For i = 1 To n
    For j = 0 To m -1
      '' Count of solutions including S[j]
      x = IIf (i >= S(j), table(i - S(j), j), 0)
      '' Count of solutions excluding S[j]
      y = IIf (j >= 1, table(i, j -1), 0)
      ''total count
      table(i, j) = x + y
    Next
  Next

  Return table(n, m -1)

End Function

' ------=< MAIN >=------

Dim As UInteger n
Dim As UInteger value()

ReDim value(3)
value(0) = 1 : value(1) = 5 : value(2) = 10 : value(3) = 25

n = 100
print
Print " There are "; count(value(), n); " ways to make change for $";n/100;" with 4 coins"
Print

n = 100000
Print " There are "; count(value(), n); " ways to make change for $";n/100;" with 4 coins"
Print

ReDim value(5)
value(0) =  1 : value(1) =  5 : value(2) =  10
value(3) = 25 : value(4) = 50 : value(5) = 100

n = 100000
Print " There are "; count(value(), n); " ways to make change for $";n/100;" with 6 coins"
Print

' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Output:
 There are 242 ways to make change for $ 1 with 4 coins

 There are 133423351001 ways to make change for $ 1000 with 4 coins

 There are 13398445413854501 ways to make change for $ 1000 with 6 coins

FutureBasic

include "NSLog.incl"

void local fn Doit
  long penny, nickel, dime, quarter, count = 0
  
  NSLogSetTabInterval(30)
  
  for penny = 0 to 100
    for nickel = 0 to 20
      for dime = 0 to 10
        for quarter = 0 to 4
          if penny + nickel * 5 + dime * 10 + quarter * 25 == 100
            NSLog(@"%ld pennies\t%ld nickels\t%ld dimes\t%ld quarters",penny,nickel,dime,quarter)
            count++
          end if
        next quarter
      next dime
    next nickel
  next penny
  
  NSLog(@"\n%ld ways to make a dollar",count)
end fn

fn DoIt

HandleEvents

Output:

0 pennies       0 nickels       0 dimes       4 quarters
0 pennies       0 nickels	5 dimes       2 quarters
0 pennies       0 nickels	10 dimes      0 quarters
0 pennies       1 nickels	2 dimes       3 quarters
......
65 pennies      5 nickels       1 dimes       0 quarters
65 pennies      7 nickels       0 dimes       0 quarters
70 pennies      0 nickels       3 dimes       0 quarters
70 pennies      1 nickels       0 dimes       1 quarters

242 ways to make a dollar

Go

Translation of: lisp
package main

import "fmt"

func main() {
    amount := 100
    fmt.Println("amount, ways to make change:", amount, countChange(amount))
}

func countChange(amount int) int64 {
    return cc(amount, 4)
}

func cc(amount, kindsOfCoins int) int64 {
    switch {
    case amount == 0:
        return 1
    case amount < 0 || kindsOfCoins == 0:
        return 0
    }
    return cc(amount, kindsOfCoins-1) +
        cc(amount - firstDenomination(kindsOfCoins), kindsOfCoins)
}

func firstDenomination(kindsOfCoins int) int {
    switch kindsOfCoins {
    case 1:
        return 1
    case 2:
        return 5
    case 3:
        return 10
    case 4:
        return 25
    }
    panic(kindsOfCoins)
}

Output:

amount, ways to make change: 100 242

Alternative algorithm, practical for the optional task.

package main

import "fmt"

func main() {
    amount := 1000 * 100
    fmt.Println("amount, ways to make change:", amount, countChange(amount))
}

func countChange(amount int) int64 {
    ways := make([]int64, amount+1)
    ways[0] = 1
    for _, coin := range []int{100, 50, 25, 10, 5, 1} {
        for j := coin; j <= amount; j++ {
            ways[j] += ways[j-coin]
        }
    }
    return ways[amount]
}

Output:

amount, ways to make change: 100000 13398445413854501

Groovy

Translation of: Go

Intuitive Recursive Solution:

def ccR
ccR = { BigInteger tot, List<BigInteger> coins ->
    BigInteger n = coins.size()
    switch ([tot:tot, coins:coins]) {
        case { it.tot == 0 } :
            return 1g
        case { it.tot < 0 || coins == [] } :
            return 0g
        default:
            return ccR(tot, coins[1..<n]) +
                ccR(tot - coins[0], coins)
    }
}

Fast Iterative Solution:

def ccI = { BigInteger tot, List<BigInteger> coins ->
    List<BigInteger> ways = [0g] * (tot+1)
    ways[0] = 1g
    coins.each { BigInteger coin ->
        (coin..tot).each { j ->
            ways[j] += ways[j-coin]
        }
    }
    ways[tot]
}

Test:

println '\nBase:'
[iterative: ccI, recursive: ccR].each { label, cc ->
    print "${label} "
    def start = System.currentTimeMillis()
    def ways = cc(100g, [25g, 10g, 5g, 1g])
    def elapsed = System.currentTimeMillis() - start
    println ("answer: ${ways}   elapsed: ${elapsed}ms")
}

print '\nExtra Credit:\niterative '
def start = System.currentTimeMillis()
def ways = ccI(1000g * 100, [100g, 50g, 25g, 10g, 5g, 1g])
def elapsed = System.currentTimeMillis() - start
println ("answer: ${ways}   elapsed: ${elapsed}ms")

Output:

Base:
iterative answer: 242   elapsed: 5ms
recursive answer: 242   elapsed: 220ms

Extra Credit:
iterative answer: 13398445413854501   elapsed: 1077ms

Haskell

Naive implementation:

count :: (Integral t, Integral a) => t -> [t] -> a
count 0 _ = 1
count _ [] = 0
count x (c:coins) =
  sum
    [ count (x - (n * c)) coins
    | n <- [0 .. (quot x c)] ]

main :: IO ()
main = print (count 100 [1, 5, 10, 25])

Much faster, probably harder to read, is to update results from bottom up:

count :: Integral a => [Int] -> [a]
count = foldr addCoin (1 : repeat 0)
  where
    addCoin c oldlist = newlist
      where
        newlist = take c oldlist ++ zipWith (+) newlist (drop c oldlist)
 
main :: IO ()
main = do
  print (count [25, 10, 5, 1] !! 100)
  print (count [100, 50, 25, 10, 5, 1] !! 10000)

Or equivalently, (reformulating slightly, and adding a further test):

import Data.Function (fix)

count
  :: Integral a
  => [Int] -> [a]
count =
  foldr
    (\x a ->
        let (l, r) = splitAt x a
        in fix ((<>) l . flip (zipWith (+)) r))
    (1 : repeat 0)

---------------------------- TEST --------------------------
main :: IO ()
main =
  mapM_
    (print . uncurry ((!!) . count))
    [ ([25, 10, 5, 1], 100)
    , ([100, 50, 25, 10, 5, 1], 10000)
    , ([100, 50, 25, 10, 5, 1], 1000000)
    ]
Output:
242
139946140451
1333983445341383545001

Icon and Unicon

procedure main()

   US_coins       := [1, 5, 10, 25]
   US_allcoins    := [1,5,10,25,50,100]
   EU_coins       := [1, 2, 5, 10, 20, 50, 100, 200]
   CDN_coins      := [1,5,10,25,100,200]
   CDN_allcoins   := [1,5,10,25,50,100,200]

   every trans := ![ [15,US_coins], 
                     [100,US_coins], 
                     [1000*100,US_allcoins] 
                  ] do 
      printf("There are %i ways to count change for %i using %s coins.\n",CountCoins!trans,trans[1],ShowList(trans[2]))
end

procedure ShowList(L)            # helper list to string 
every (s := "[ ") ||:= !L || " "
return s || "]"
end

This is a naive implementation and very slow.

This example is in need of improvement:

Needs a better algorithm.

procedure CountCoins(amt,coins)  # very slow, recurse by coin value
local count
static S

if type(coins) == "list" then {
   S := sort(set(coins))
   if *S < 1 then runerr(205,coins)
   return  CountCoins(amt)
   }
else {
   /coins := 1
   if value := S[coins] then {
      every (count := 0) +:= CountCoins(amt - (0 to amt by value), coins + 1) 
      return count
      }   
   else    
      return (amt ~= 0) | 1
   }
end

printf.icn provides formatting

Output:

There are 6 ways to count change for 15 using [ 1 5 10 25 ] coins.
There are 242 ways to count change for 100 using [ 1 5 10 25 ] coins.
^c

Another one:

# coin.icn
# usage: coin value
procedure count(coinlist, value)
	if value = 0 then return 1
	if value < 0 then return 0
	if (*coinlist <= 0) & (value >= 1) then return 0
	return count(coinlist[1:*coinlist], value) + count(coinlist, value - coinlist[*coinlist])
end


procedure main(params)
	money := params[1]
	coins := [1,5,10,25]
	
	writes("Value of ", money, " can be changed by using a set of ")
	every writes(coins[1 to *coins], " ")
	write(" coins in ", count(coins, money), " different ways.")
end

Output:

Value of 15 can be changed by using a set of 1 5 10 25  coins in 6 different ways.
Value of 100 can be changed by using a set of 1 5 10 25  coins in 242 different ways.

IS-BASIC

100 PROGRAM "Coins.bas"
110 LET MONEY=100
120 LET COUNT=0
125 PRINT "Count  Pennies Nickles Dimes Quaters"
130 FOR QC=0 TO INT(MONEY/25)
150   FOR DC=0 TO INT((MONEY-QC*25)/10)
170     FOR NC=0 TO INT((MONEY-DC*10)/5)
190       FOR PC=0 TO MONEY-NC*5 STEP 5
200         LET S=PC+NC*5+DC*10+QC*25
210         IF S=MONEY THEN
220           LET COUNT=COUNT+1
230           PRINT COUNT,PC,NC,DC,QC
240         END IF
250       NEXT
260     NEXT
270   NEXT
280 NEXT
290 PRINT COUNT;"different combinations found."

J

In this draft intermediate results are a two column array. The first column is tallies -- the number of ways we have for reaching the total represented in the second column, which is unallocated value (which we will assume are pennies). We will have one row for each different in-range value which can be represented using only nickles (0, 5, 10, ... 95, 100).

merge=: ({:"1 (+/@:({."1),{:@{:)/. ])@;
count=: {.@] <@,. {:@] - [ * [ i.@>:@<.@%~ {:@]
init=: (1 ,. ,.)^:(0=#@$)
nsplits=: 0 { [: +/ [: (merge@:(count"1) init)/ }.@/:~@~.@,

This implementation special cases the handling of pennies and assumes that the lowest coin value in the argument is 1. If I needed additional performance, I would next special case the handling of nickles/penny combinations...

Thus:

   100 nsplits 1 5 10 25
242

And, on a 64 bit machine with sufficient memory:

   100000 nsplits 1 5 10 25 50 100
13398445413854501

Warning: the above version can miss one when the largest coin is equal to the total value.

For British viewers change from £10 using £10 £5 £2 £1 50p 20p 10p 5p 2p and 1p

   init =: 4 : '(1+x)$1'
length1 =: 4 : '1=#y'
      f =: 4 : ',/ +/\ (-x) ]\ y'

      1000 {  f ` init @. length1 / 1000 500 200 100 50 20 10 5 2 , 1000 0
327631322

NB. this is a foldLeft once initialised the intermediate right arguments are arrays
 1000 f 500 f 200 f 100 f 50 f 20 f 10 f 5 f 2 f (1000 init 0)

Java

Translation of: D
Works with: Java version 1.5+
import java.util.Arrays;
import java.math.BigInteger;

class CountTheCoins {
    private static BigInteger countChanges(int amount, int[] coins){
        final int n = coins.length;
        int cycle = 0;
        for (int c : coins)
            if (c <= amount && c >= cycle)
                cycle = c + 1;
        cycle *= n;
        BigInteger[] table = new BigInteger[cycle];
        Arrays.fill(table, 0, n, BigInteger.ONE);
        Arrays.fill(table, n, cycle, BigInteger.ZERO);

        int pos = n;
        for (int s = 1; s <= amount; s++) {
            for (int i = 0; i < n; i++) {
                if (i == 0 && pos >= cycle)
                    pos = 0;
                if (coins[i] <= s) {
                    final int q = pos - (coins[i] * n);
                    table[pos] = (q >= 0) ? table[q] : table[q + cycle];
                }
                if (i != 0)
                    table[pos] = table[pos].add(table[pos - 1]);
                pos++;
            }
        }

        return table[pos - 1];
    }

    public static void main(String[] args) {
        final int[][] coinsUsEu = {{100, 50, 25, 10, 5, 1},
                                   {200, 100, 50, 20, 10, 5, 2, 1}};

        for (int[] coins : coinsUsEu) {
            System.out.println(countChanges(     100,
                Arrays.copyOfRange(coins, 2, coins.length)));
            System.out.println(countChanges(  100000, coins));
            System.out.println(countChanges( 1000000, coins));
            System.out.println(countChanges(10000000, coins) + "\n");
        }
    }
}

Output:

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

JavaScript

Iterative

Efficient iterative algorithm (cleverly calculates number of combinations without permuting them)

function countcoins(t, o) {
    'use strict';
    var targetsLength = t + 1;
    var operandsLength = o.length;
    t = [1];

    for (var a = 0; a < operandsLength; a++) {
        for (var b = 1; b < targetsLength; b++) {

            // initialise undefined target
            t[b] = t[b] ? t[b] : 0;

            // accumulate target + operand ways
            t[b] += (b < o[a]) ? 0 : t[b - o[a]];
        }
    }

    return t[targetsLength - 1];
}
Output:

JavaScript hits integer limit for optional task

countcoins(100, [1,5,10,25]);
242

Recursive

Inefficient recursive algorithm (naively calculates number of combinations by actually permuting them)

function countcoins(t, o) {
    'use strict';
    var operandsLength = o.length;
    var solutions = 0;

    function permutate(a, x) {

        // base case
        if (a === t) {
            solutions++;
        }

        // recursive case
        else if (a < t) {
            for (var i = 0; i < operandsLength; i++) {
                if (i >= x) {
                    permutate(o[i] + a, i);
                }
            }
        }
    }

    permutate(0, 0);
    return solutions;
}
Output:

Too slow for optional task

countcoins(100, [1,5,10,25]);
242

Iterative again

Translation of: C#
var amount = 100,
    coin = [1, 5, 10, 25]
var t = [1];
for (t[amount] = 0, a = 1; a < amount; a++) t[a] = 0 // initialise t[0..amount]=[1,0,...,0]
for (var i = 0, e = coin.length; i < e; i++)
    for (var ci = coin[i], a = ci; a <= amount; a++)
        t[a] += t[a - ci]
document.write(t[amount])
Output:
242

jq

Currently jq uses IEEE 754 64-bit numbers. Large integers are approximated by floats, and therefore the answer that the following program provides for the optional task is only correct for the first 15 digits.

# How many ways are there to make "target" cents, given a list of coin
# denominations as input.
# The strategy is to record at total[n] the number of ways to make n cents.
def countcoins(target):
  . as $coin
  | reduce range(0; length) as $a
      ( [1];   # there is 1 way to make 0 cents
        reduce range(1; target + 1) as $b
          (.;                                      # total[]
           if $b < $coin[$a] then .
           else  .[$b - $coin[$a]] as $count
           | if $count == 0 then .
             else .[$b] += $count
             end
           end ) ) 
  | .[target] ;

Example:

[1,5,10,25] | countcoins(100)
Output:
242

Julia

Translation of: Python
function changes(amount::Int, coins::Array{Int})::Int128
    ways = zeros(Int128, amount + 1)
    ways[1] = 1
    for coin in coins, j in coin+1:amount+1
        ways[j] += ways[j - coin]
    end
    return ways[amount + 1]
end

@show changes(100, [1, 5, 10, 25])
@show changes(100000, [1, 5, 10, 25, 50, 100])
Output:
changes(100, [1, 5, 10, 25]) = 242
changes(100000, [1, 5, 10, 25, 50, 100]) = 13398445413854501

Kotlin

Translation of: C#
// version 1.0.6

fun countCoins(c: IntArray, m: Int, n: Int): Long {
    val table = LongArray(n + 1)
    table[0] = 1
    for (i in 0 until m) 
        for (j in c[i]..n) table[j] += table[j - c[i]]
    return table[n]
}

fun main(args: Array<String>) {
    val c = intArrayOf(1, 5, 10, 25, 50, 100)
    println(countCoins(c, 4, 100))
    println(countCoins(c, 6, 1000 * 100)) 
}
Output:
242
13398445413854501

Lasso

Inspired by the javascript iterative example for the same task

define cointcoins(
	target::integer,
	operands::array
) => {

	local(
		targetlength	= #target + 1,
		operandlength	= #operands -> size,
		output			= staticarray_join(#targetlength,0),
		outerloopcount
	)

	#output -> get(1) = 1

	loop(#operandlength) => {
		#outerloopcount = loop_count
		loop(#targetlength) => {

			if(loop_count >= #operands -> get(#outerloopcount) and loop_count - #operands -> get(#outerloopcount) > 0) => {
				#output -> get(loop_count) += #output -> get(loop_count - #operands -> get(#outerloopcount))
			}
		}
	}

	return #output -> get(#targetlength)
}

cointcoins(100, array(1,5,10,25,))
'<br />'
cointcoins(100000, array(1, 5, 10, 25, 50, 100))

Output:

242
13398445413854501

Lua

Lua uses one-based indexes but table keys can be any value so you can define an element 0 just as easily as you can define an element "foo"...

function countSums (amount, values)
    local t = {}
    for i = 1, amount do t[i] = 0 end
    t[0] = 1
    for k, val in pairs(values) do
        for i = val, amount do t[i] = t[i] + t[i - val] end
    end
    return t[amount]
end

print(countSums(100, {1, 5, 10, 25}))
print(countSums(100000, {1, 5, 10, 25, 50, 100}))
Output:
242
1.3398445413855e+16

M2000 Interpreter

Fast O(n*m)

Works with decimals in table()

Module FindCoins {
      Function count(c(), n)  {
            dim table(n+1)=0@ :  table(0)=1@
            for c=0 to len(c())-1 {
                 if c(c)>n then exit
            }
            if c else exit
            for i=0 to c-1 {for j=c(i) to n {table(j)+=table(j-c(i))}}
            =table(n)
      }
      Print "For 1$ ways to change:";count((1,5,10,25),100)
      Print "For 100$ (optional task ways to change):";count((1,5,10,25,50,100),100000)
}
FindCoins
Output:
For 1$ ways to change:242
For 100$ (optional task) ways to change:13398445413854501

With Recursion with saving partial results

Using an inventory (a kind of vector) to save first search (but is slower than previous one)

Module CheckThisToo {
      inventory c=" 0 0":=1@
      make_change=lambda c (amount, coins()) ->{
            m=lambda c,coins() (n,m)->{if n<0 or m<0 then =0@:exit
            if exist(c,str$(n)+str$(m)) then =eval(c):exit
            append c,str$(n)+str$(m):=lambda(n-coins(m), m)+lambda(n, m-1):=c(str$(n)+str$(m))}
           =m(amount,len(coins())-1)
      }
      Print make_change(100, (1,5,10,25,50,100))=293
      Print make_change(100, (1,5,10,25))=242
      Print make_change(15, (1,5,10,25))=6
      Print make_change(5, (1,5,10,25))=2
}
CheckThisToo

MAD

            NORMAL MODE IS INTEGER
            DIMENSION TAB(101)

            THROUGH ZERO, FOR N = 1, 1, N.G.100
ZERO        TAB(N) = 0
            TAB(0) = 1

            THROUGH STEP, FOR VALUES OF COIN = 1, 5, 10, 25
            THROUGH STEP, FOR N = COIN, 1, N.G.100
STEP        TAB(N) = TAB(N) + TAB(N - COIN)

            VECTOR VALUES FMT = $I3*$
            PRINT FORMAT FMT, TAB(100)
            END OF PROGRAM
Output:
242

Maple

Straightforward implementation with power series. Not very efficient for large amounts. Note that in the following, all amounts are in cents.

assume(p::posint,abs(x)<1):
coin:=unapply(sum(x^(p*n),n=0..infinity),p):
ways:=(amount,purse)->coeff(series(mul(coin(k),k in purse),x,amount+1),x,amount):

ways(100,[1,5,10,25]);
# 242

ways(1000,[1,5,10,25,50,100]);
# 2103596

ways(10000,[1,5,10,25,50,100]);
# 139946140451

ways(100000,[1,5,10,25,50,100]);
# 13398445413854501

A faster implementation.

ways2:=proc(amount,purse)
  local a,n,k;
  a:=Array(1..amount);
  for k in purse do
    for n from k to amount do
      if n=k then
        a[n]++;
      else
        a[n]+=a[n-k]
      fi
    od
  od;
  a[-1]
end:

ways2(100,[1,5,10,25]);
# 242

ways2(100,[1,5,10,25,50,100]);
# 293

ways2(1000,[1,5,10,25,50,100]);
# 2103596

ways2(10000,[1,5,10,25,50,100]);
# 139946140451

ways2(100000,[1,5,10,25,50,100]);
# 13398445413854501

ways2(1000000,[1,5,10,25,50,100]);
# 1333983445341383545001

ways2(10000000,[1,5,10,25,50,100]);
# 133339833445334138335450001

ways2(100000000,[1,5,10,25,50,100]);
# 13333398333445333413833354500001

Additionally, while it's not proved as is, we can see that the first values for an amount 10^k obey the following simple formula:

P:=n->4/(3*10^9)*n^5+65/10^8*n^4+112/10^6*n^3+805/10^5*n^2+635/3000*n+1:

for k from 2 to 8 do lprint(P(10^k)) od:
293
2103596
139946140451
13398445413854501
1333983445341383545001
133339833445334138335450001
13333398333445333413833354500001

The polynomial P(n) seems to give the correct number of ways iff n is a multiple of 100 (tested up to n=10000000), i.e. the number of ways for 100n is

Q:=n->40/3*n^5+65*n^4+112*n^3+161/2*n^2+127/6*n+1:

Mathematica / Wolfram Language

Translation of: Go
CountCoins[amount_, coinlist_] := ( ways = ConstantArray[1, amount];
Do[For[j = coin, j <= amount, j++,
  If[ j - coin == 0,
    ways[[j]] ++,
    ways[[j]] += ways[[j - coin]]
]]
, {coin, coinlist}];
ways[[amount]])

Example usage:

CountCoins[100, {25, 10, 5}]
-> 242

CountCoins[100000, {100, 50, 25, 10, 5}]
-> 13398445413854501

MATLAB / Octave

%% Count_The_Coins
clear;close all;clc;
tic
 
for i = 1:2 % 1st loop is main challenge 2nd loop is optional challenge
    if (i == 1)
        amount = 100;                       % Matlab indexes from 1 not 0, so we need to add 1 to our target value                        
        amount = amount + 1;                    
        coins = [1 5 10 25];                % Value of coins we can use
    else
        amount = 100*1000;                  % Matlab indexes from 1 not 0, so we need to add 1 to our target value                        
        amount = amount + 1; 
        coins = [1 5 10 25 50 100];         % Value of coins we can use
    end % End if
    ways = zeros(1,amount);                 % Preallocating for speed
    ways(1) = 1;                            % First solution is 1
 
    % Solves from smallest sub problem to largest (bottom up approach of dynamic programming).
    for j = 1:length(coins)                 
        for K = coins(j)+1:amount           
            ways(K) = ways(K) + ways(K-coins(j));   
        end % End for
    end % End for
        if (i == 1)
            fprintf(Main Challenge: %d \n', ways(amount));
        else
            fprintf(Bonus Challenge: %d \n', ways(amount));
        end % End if 
end % End for
toc

Example Output:

Main Challenge: 242

Bonus Challenge: 13398445413854501

Mercury

:- module coins.
:- interface.
:- import_module int, io.
:- type coin ---> quarter; dime; nickel; penny.
:- type purse ---> purse(int, int, int, int).

:- pred sum_to(int::in, purse::out) is nondet.

:- pred main(io::di, io::uo) is det.
:- implementation.
:- import_module solutions, list, string.

:- func value(coin) = int.
value(quarter) = 25.
value(dime) = 10.
value(nickel) = 5.
value(penny) = 1.

:- pred supply(coin::in, int::in, int::out) is multi.
supply(C, Target, N) :- upto(Target div value(C), N).

:- pred upto(int::in, int::out) is multi.
upto(N, R) :- ( nondet_int_in_range(0, N, R0) -> R = R0 ; R = 0 ).

sum_to(To, Purse) :-
	Purse = purse(Q, D, N, P),
	sum(Purse) = To,
	supply(quarter, To, Q),
	supply(dime, To, D),
	supply(nickel, To, N),
	supply(penny, To, P).

:- func sum(purse) = int.
sum(purse(Q, D, N, P)) =
	value(quarter) * Q + value(dime) * D +
	value(nickel) * N + value(penny) * P.

main(!IO) :-
	solutions(sum_to(100), L),
	show(L, !IO),
	io.format("There are %d ways to make change for a dollar.\n",
                  [i(length(L))], !IO).

:- pred show(list(purse)::in, io::di, io::uo) is det.
show([], !IO).
show([P|T], !IO) :-
	io.write(P, !IO), io.nl(!IO),
	show(T, !IO).

Nim

Translation of: Python
proc changes(amount: int, coins: openArray[int]): int =
  var ways = @[1]
  ways.setLen(amount+1)
  for coin in coins:
    for j in coin..amount:
      ways[j] += ways[j-coin]
  ways[amount]

echo changes(100, [1, 5, 10, 25])
echo changes(100000, [1, 5, 10, 25, 50, 100])

Output:

242
13398445413854501

OCaml

Translation of the D minimal version:

let changes amount coins =
  let ways = Array.make (amount + 1) 0L in
  ways.(0) <- 1L;
  List.iter (fun coin ->
    for j = coin to amount do
      ways.(j) <- Int64.add ways.(j) ways.(j - coin)
    done
  ) coins;
  ways.(amount)

let () =
  Printf.printf "%Ld\n" (changes    1_00 [25; 10; 5; 1]);
  Printf.printf "%Ld\n" (changes 1000_00 [100; 50; 25; 10; 5; 1]);
;;

Output:

$ ocaml coins.ml 
242
13398445413854501

PARI/GP

coins(v)=prod(i=1,#v,1/(1-'x^v[i]));
ways(v,n)=polcoeff(coins(v)+O('x^(n+1)),n);
ways([1,5,10,25],100)
ways([1,5,10,25,50,100],100000)

Output:

%1 = 242
%2 = 13398445413854501

Pascal

program countTheCoins;

{$mode objfpc}{$H+}

var
  count, quarter, dime, nickel, penny: integer;

begin
  count := 0;

  for penny := 0 to 100 do
    for nickel := 0 to 20 do
      for dime := 0 to 10 do
        for quarter := 0 to 4 do
          if (penny + 5 * nickel + 10 * dime + 25 * quarter = 100) then
          begin
            writeln(penny, ' pennies ', nickel, ' nickels ', dime, ' dimes ', quarter, ' quarters');
            count := count + 1;
          end;


  writeln('The number of ways to make change for a dollar is: ', count); // 242 ways to make change for a dollar

end.

Output:

0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
0 pennies 1 nickels 2 dimes 3 quarters
......
85 pennies 1 nickels 1 dimes 0 quarters
85 pennies 3 nickels 0 dimes 0 quarters
90 pennies 0 nickels 1 dimes 0 quarters
90 pennies 2 nickels 0 dimes 0 quarters
95 pennies 1 nickels 0 dimes 0 quarters
100 pennies 0 nickels 0 dimes 0 quarters

The number of ways to make change for a dollar is: 242

Perl

use 5.01;
use Memoize;

sub cc {
    my $amount = shift;
    return 0 if !@_ || $amount < 0;
    return 1 if $amount == 0;
    my $first = shift;
    cc( $amount, @_ ) + cc( $amount - $first, $first, @_ );
}
memoize 'cc';

# Make recursive algorithm run faster by sorting coins descending by value:
sub cc_optimized {
    my $amount = shift;
    cc( $amount, sort { $b <=> $a } @_ );
}

say 'Ways to change $ 1 with common coins: ',
    cc_optimized( 100, 1, 5, 10, 25 );
say 'Ways to change $ 1000 with addition of less common coins: ',
    cc_optimized( 1000 * 100, 1, 5, 10, 25, 50, 100 );
Output:
Ways to change $ 1 with common coins: 242
Ways to change $ 1000 with addition of less common coins: 13398445413854501

Phix

Very fast, from http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change

function coin_count(sequence coins, integer amount)
    sequence s = repeat(0,amount+1)
    s[1] = 1
    for c=1 to length(coins) do
        for n=coins[c] to amount do
            s[n+1] += s[n-coins[c]+1]
        end for
    end for
    return s[amount+1]
end function

An attempt to explain this algorithm further seems worthwhile:

function coin_count(sequence coins, integer amount)
    -- start with 1 known way to achieve 0 (being no coins)
    --  (nb: s[1] holds the solution for 0, s[n+1] for n)
    sequence s = repeat(0,amount+1)
    s[1] = 1
    -- then for every coin that we can use, increase number of 
    --  solutions by that previously found for the remainder.
    for c=1 to length(coins) do
        -- this inner loop is essentially behaving as if we had
        -- called this routine with 1..amount, but skipping any 
        -- less than the coin's value, hence coins[c]..amount.
        for n=coins[c] to amount do
            s[n+1] += s[n-coins[c]+1]
        end for
    end for
    return s[amount+1]
end function
 
-- The key to understanding the above is to try a dry run of this:
printf(1,"%d\n",coin_count({2,3},5))    -- (prints 1)
-- You'll need 4 2p coins, 3 3p coins, and 5 spaces marked 1..5.
-- Place 2p wherever it fits: 1:0 2:1 3:1 4:1 5:1
-- Add previously found solns: +0  +1  +0  +1  +0   [1]
-- Place 3p wherever it fits: 1:0 2:0 3:1 4:1 5:1
-- Add previously found solns: +0  +0  +1  +0  +1   [2]
-- [1] obviously at 2: we added the base soln for amount=0,
--     and at 4: we added the previously found soln for 2.
--     also note that we added nothing for 2p+3p, yet, that
--     fact is central to understanding why this works. [3]
-- [2] obviously at 3: we added the base soln for amount=0,
--     at 4: we added the zero solutions yet found for 1p,
--     and at 5: we added the previously found soln for 2.
--     you can imagine at 6,9,12 etc all add in soln for 3,
--     albeit by adding that as just added to the precessor.
-- [3] since we add no 3p solns when processing 2p, we do 
--     not count 2p+3p and 3p+2p as two solutions.
 
--For N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}.
printf(1,"%d\n",coin_count({1,2,3},4))
--For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}.
printf(1,"%d\n\n",coin_count({2,3,5,6},10))
 
printf(1,"%d\n",coin_count({25, 10, 5, 1},1_00))
printf(1,"%,d\n",coin_count({100, 50, 25, 10, 5, 1},1000_00))
Output:
1
4
5

242
13,398,445,413,854,501

Note that a slightly wrong value is printed when running this on 32 bits:

13,398,445,413,854,501      -- 64 bit (exact)
13,398,445,413,854,496      -- 32 bit (5 out)
 9,007,199,254,740,992      -- max precision (53 bits) of a 64-bit float

Picat

Using dynamic programming with tabling.

go =>
  Problems = [[      1*100, [25,10,5,1]],            % 1 dollar
              [    100*100, [100,50,25,10,5,1]],   % 100 dollars
              [  1_000*100, [100,50,25,10,5,1]],  % 1000 dollars
              [ 10_000*100, [100,50,25,10,5,1]], % 10000 dollars
              [100_000*100, [100,50,25,10,5,1]] % 100000 dollars
              ],
  foreach([N,L] in Problems)
     initialize_table, % clear the tabling from previous run
     println([n=N,l=L]),
     time(println(num_sols=coins(L,N,1)))
  end.

table
coins(Coins, Money, M) = Sum =>
    Sum1 = 0,
    Len = Coins.length,
    if M == Len then
      Sum1 := 1,
    else 
       foreach(I in M..Len)
         if Money - Coins[I] == 0 then
            Sum1 := Sum1 + 1
         end,
         if Money - Coins[I] > 0 then
            Sum1 := Sum1 + coins(Coins, Money-Coins[I], I)
         end,
       end
    end,
    Sum = Sum1.
Output:
[n = 100,l = [25,10,5,1]]
num_sols = 242

CPU time 0.0 seconds.

[n = 10000,l = [100,50,25,10,5,1]]
num_sols = 139946140451

CPU time 0.005 seconds.

[n = 100000,l = [100,50,25,10,5,1]]
num_sols = 13398445413854501

CPU time 0.046 seconds.

[n = 1000000,l = [100,50,25,10,5,1]]
num_sols = 1333983445341383545001

CPU time 0.496 seconds.

[n = 10000000,l = [100,50,25,10,5,1]]
num_sols = 133339833445334138335450001

CPU time 5.402 seconds.

PicoLisp

Translation of: C
(de coins (Sum Coins)
   (let (Buf (mapcar '((N) (cons 1 (need (dec N) 0))) Coins)  Prev)
      (do Sum
         (zero Prev)
         (for L Buf
            (inc (rot L) Prev)
            (setq Prev (car L)) ) )
      Prev ) )

Test:

(for Coins '((100 50 25 10 5 1) (200 100 50 20 10 5 2 1))
   (println (coins 100 (cddr Coins)))
   (println (coins (* 1000 100) Coins))
   (println (coins (* 10000 100) Coins))
   (println (coins (* 100000 100) Coins))
   (prinl) )

Output:

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

Prolog

Basic version using brute force and constraint programming, the bonus version will work but takes a long time so skipped it.

:- use_module(library(clpfd)).

% Basic, Q = Quarter, D = Dime, N = Nickel, P = Penny
coins(Q, D, N, P, T) :-
	[Q,D,N,P] ins 0..T,
	T #= (Q * 25) + (D * 10) + (N * 5) + P.

coins_for(T) :-
	coins(Q,D,N,P,T),
	maplist(indomain, [Q,D,N,P]).
Output:
?- aggregate(count, coins_for(100), Count).
Count = 242.

Python

Simple version

Translation of: Go
def changes(amount, coins):
    ways = [0] * (amount + 1)
    ways[0] = 1
    for coin in coins:
        for j in xrange(coin, amount + 1):
            ways[j] += ways[j - coin]
    return ways[amount]

print changes(100, [1, 5, 10, 25])
print changes(100000, [1, 5, 10, 25, 50, 100])

Output:

242
13398445413854501

Fast version

Translation of: C
try:
    import psyco
    psyco.full()
except ImportError:
    pass

def count_changes(amount_cents, coins):
    n = len(coins)
    # max([]) instead of max() for Psyco
    cycle = max([c+1 for c in coins if c <= amount_cents]) * n
    table = [0] * cycle
    for i in xrange(n):
        table[i] = 1

    pos = n
    for s in xrange(1, amount_cents + 1):
        for i in xrange(n):
            if i == 0 and pos >= cycle:
                pos = 0
            if coins[i] <= s:
                q = pos - coins[i] * n
                table[pos]= table[q] if (q >= 0) else table[q + cycle]
            if i:
                table[pos] += table[pos - 1]
            pos += 1
    return table[pos - 1]

def main():
    us_coins = [100, 50, 25, 10, 5, 1]
    eu_coins = [200, 100, 50, 20, 10, 5, 2, 1]

    for coins in (us_coins, eu_coins):
        print count_changes(     100, coins[2:])
        print count_changes(  100000, coins)
        print count_changes( 1000000, coins)
        print count_changes(10000000, coins), "\n"

main()

Output:

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

Quackery

  [ stack ]                      is lim        (     --> s )

  [ swap dup 1+ lim put
    1 0 rot of join
    swap witheach
      [ 0 over of
        swap negate temp put
        lim share times
          [ over i^ peek
            over temp share peek
            + join ]
        temp take negate split
        nip nip ]
    -1 peek
    lim release ]                is makechange ( n [ --> n )

  say "With US coins." cr
  100 ' [ 1 5 10 25 ] makechange echo cr
  100000 ' [ 1 5 10 25 50 100 ] makechange echo cr
  cr
  say "With EU coins." cr
  100 ' [ 1 2 5 10 20 50 100 200 ] makechange echo cr
  100000 ' [ 1 2 5 10 20 50 100 200 ] makechange echo cr
Output:
With US coins.
242
13398445413854501

With EU coins.
4563
10056050940818192726001

Racket

This is the basic recursive way:

#lang racket
(define (ways-to-make-change cents coins)
  (cond ((null? coins) 0)
        ((negative? cents) 0)
        ((zero? cents) 1)
        (else
         (+ (ways-to-make-change cents (cdr coins))
            (ways-to-make-change (- cents (car coins)) coins)))))

(ways-to-make-change 100 '(25 10 5 1)) ; -> 242

This works for the small numbers, but the optional task is just too slow with this solution, so with little change to the code we can use memoization:

#lang racket
 
(define memos (make-hash))
(define (ways-to-make-change cents coins)
  (cond [(or (empty? coins) (negative? cents)) 0]
        [(zero? cents) 1]
        [else (define (answerer-for-new-arguments)
                (+ (ways-to-make-change cents (rest coins))
                   (ways-to-make-change (- cents (first coins)) coins)))
              (hash-ref! memos (cons cents coins) answerer-for-new-arguments)]))

(time (ways-to-make-change 100 '(25 10 5 1)))
(time (ways-to-make-change 100000 '(100 50 25 10 5 1)))
(time (ways-to-make-change 1000000 '(200 100 50 20 10 5 2 1)))

#| Times in milliseconds, and results:

     cpu time: 1 real time: 1 gc time: 0
     242

     cpu time: 524 real time: 553 gc time: 163
     13398445413854501

     cpu time: 20223 real time: 20673 gc time: 10233
     99341140660285639188927260001 |#

Raku

(formerly Perl 6)

Works with: rakudo version 2018.10
Translation of: Ruby
# Recursive (cached)
sub change-r($amount, @coins) {
    my @cache = [1 xx @coins], |([] xx $amount);

    multi ways($n where $n >= 0, @now [$coin,*@later]) {
        @cache[$n;+@later] //= ways($n - $coin, @now) + ways($n, @later);
    }
    multi ways($,@) { 0 }

    # more efficient to start with coins sorted in descending order
    ways($amount, @coins.sort(-*).list);
}

# Iterative
sub change-i(\n, @coins) {
    my @table = [1 xx @coins], [0 xx @coins] xx n;
    (1..n).map: -> \i {
        for ^@coins -> \j {
        my \c = @coins[j];
        @table[i;j] = [+]
            @table[i - c;j] // 0,
            @table[i;j - 1] // 0;
        }
    }
    @table[*-1][*-1];
}

say "Iterative:";
say change-i    1_00, [1,5,10,25];
say change-i 1000_00, [1,5,10,25,50,100];

say "\nRecursive:";
say change-r    1_00, [1,5,10,25];
say change-r 1000_00, [1,5,10,25,50,100];
Output:
Iterative:
242
13398445413854501

Recursive:
242
13398445413854501

REXX

recursive

The recursive calls to the subroutine have been unrolled somewhat, this reduces the number of recursive calls substantially.

These REXX versions also support fractional cents (as in a   ½-cent   and   ¼-cent coins).   Any fractional coin can be
specified as a decimal fraction   (.5,    .25,   ···).

Support was included to allow specification of half-cent and quarter-cent coins as   1/2   and   1/4.

The amount can be specified in cents (as a number), or in dollars (as for instance,   $1000).

/*REXX program counts the number of ways to make change with coins from an given amount.*/
numeric digits 20                                /*be able to handle large amounts of $.*/
parse arg N $                                    /*obtain optional arguments from the CL*/
if N='' | N=","     then N= 100                  /*Not specified?  Then Use $1  (≡100¢).*/
if $='' | $=","     then $= 1 5 10 25            /*Use penny/nickel/dime/quarter default*/
if left(N, 1)=='$'  then N= 100 * substr(N, 2)   /*the count was specified in  dollars. */
coins= words($)                                  /*the number of coins specified.       */
NN= N;              do j=1  for coins            /*create a fast way of accessing specie*/
                    _= word($, j)                /*define an array element for the coin.*/
                    if _=='1/2'  then _=.5       /*an alternate spelling of a half-cent.*/
                    if _=='1/4'  then _=.25      /* "     "         "     " " quarter-¢.*/
                    $.j= _                       /*assign the value to a particular coin*/
                    end   /*j*/
_= n//100;                       cnt=' cents'    /* [↓]  is the amount in whole dollars?*/
if _=0  then do;    NN= '$' || (NN%100);   cnt=  /*show the amount in dollars, not cents*/
             end                                 /*show the amount in dollars, not cents*/
say 'with an amount of '      comma(NN)cnt",  there are "         comma( MKchg(N, coins) )
say 'ways to make change with coins of the following denominations: '    $
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
comma: procedure; parse arg _;     n= _'.9';      #= 123456789;      b= verify(n, #, "M")
       e= verify(n, #'0', , verify(n, #"0.", 'M'))  -  4
            do j=e  to b  by -3;   _= insert(',', _, j);    end  /*j*/;          return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
MKchg: procedure expose $.;        parse arg a,k /*this function is invoked recursively.*/
       if a==0    then return 1                  /*unroll for a special case of  zero.  */
       if k==1    then return 1                  /*   "    "  "    "      "   "  unity. */
       if k==2    then f= 1                      /*handle this special case   of  two.  */
                  else f= MKchg(a, k-1)          /*count,  and then recurse the amount. */
       if a==$.k  then return f+1                /*handle this special case of A=a coin.*/
       if a <$.k  then return f                  /*   "     "     "      "   " A<a coin.*/
                       return f+MKchg(a-$.k,k)   /*use diminished amount ($) for change.*/
output   when using the default input:
with an amount of  $1,  there are  242
ways to make change with coins of the following denominations:  1 5 10 25
output   when using the following input:     $1   1/4   1/2   1   2   3   5   10   20   25   50   100
with an amount of  $1,  there are  29,034,171
ways to make change with coins of the following denominations:  1/4 1/2 1 2 3 5 10 20 25 50 100

with memoization

This REXX version is more than a couple of orders of magnitude faster than the 1st version when using larger amounts.

/*REXX program counts the number of ways to make change with coins from an given amount.*/
numeric digits 20                                /*be able to handle large amounts of $.*/
parse arg N $                                    /*obtain optional arguments from the CL*/
if N='' | N=","    then N= 100                   /*Not specified?  Then Use $1  (≡100¢).*/
if $='' | $=","    then $= 1 5 10 25             /*Use penny/nickel/dime/quarter default*/
if left(N,1)=='$'  then N= 100 * substr(N, 2)    /*the amount was specified in  dollars.*/
NN= N;                     coins= words($)       /*the number of coins specified.       */
!.= .;        do j=1  for coins                  /*create a fast way of accessing specie*/
              _= word($, j);    ?= _ ' coin'     /*define an array element for the coin.*/
              if _=='½' | _=="1/2"   then _= .5  /*an alternate spelling of a half─cent.*/
              if _=='¼' | _=="1/4"   then _= .25 /* "     "         "     " " quarter─¢.*/
              $.j= _                             /*assign the value to a particular coin*/
              end   /*j*/
_= n // 100;                    cnt=' cents'     /* [↓]  is the amount in whole dollars?*/
if _=0  then do;   NN= '$'  ||  (NN%100);  cnt=  /*show the amount in dollars, not cents*/
             end
say 'with an amount of '      comma(NN)cnt",  there are "         comma( MKchg(N, coins) )
say 'ways to make change with coins of the following denominations: '    $
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
comma: procedure; parse arg _;     n= _'.9';      #= 123456789;      b= verify(n, #, "M")
       e= verify(n, #'0', , verify(n, #"0.", 'M'))  -  4
            do j=e  to b  by -3;   _= insert(',', _, j);    end  /*j*/;          return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
MKchg: procedure expose $. !.;           parse arg a,k       /*function is recursive.   */
       if !.a.k\==.  then return !.a.k                       /*found this A & K before? */
       if a==0       then return 1                           /*unroll for a special case*/
       if k==1       then return 1                           /*   "    "  "    "      " */
       if k==2  then f= 1                                    /*handle this special case.*/
                else f= MKchg(a, k-1)                        /*count, recurse the amount*/
       if a==$.k  then do;  !.a.k= f+1;  return !.a.k;  end  /*handle this special case.*/
       if a <$.k  then do;  !.a.k= f  ;  return f    ;  end  /*   "     "     "      "  */
       !.a.k= f + MKchg(a-$.k, k);       return !.a.k        /*compute, define, return. */
output   when using the following input for the optional test case:     $1000   1   5   10   25   50   100
with an amount of  $1,000,  there are  13,398,445,413,854,501
ways to make change with coins of the following denominations:  1 5 10 25 50 100

with error checking

This REXX version is identical to the previous REXX version, but has error checking for the amount and the coins specified.

/*REXX program counts the number of ways to make change with coins from an given amount.*/
numeric digits 20                                /*be able to handle large amounts of $.*/
parse arg N $                                    /*obtain optional arguments from the CL*/
if N='' | N=","    then N= 100                   /*Not specified?  Then Use $1  (≡100¢).*/
if $='' | $=","    then $= 1 5 10 25             /*Use penny/nickel/dime/quarter default*/
X= N                                             /*save original for possible error msgs*/
if left(N,1)=='$'  then do                       /*the amount has a leading dollar sign.*/
                        _= substr(N, 2)          /*the amount was specified in  dollars.*/
                        if \isNum(_)  then call ser  "amount isn't numeric: "   N
                        N= 100 * _               /*change amount (in $) ───►  cents (¢).*/
                        end
max$= 10 ** digits()                             /*the maximum amount this pgm can have.*/
if \isNum(N)  then call  ser  X   " amount isn't numeric."
if N=0        then call  ser  X   " amount can't be zero."
if N<0        then call  ser  X   " amount can't be negative."
if N>max$     then call  ser  X   " amount can't be greater than " max$'.'
coins= words($);   !.= .;     NN= N;       p= 0  /*#coins specified; coins; amount; prev*/
@.= 0                                            /*verify a coin was only specified once*/
          do j=1  for coins;     _= word($, j)   /*create a fast way of accessing specie*/
          ?= _  ' coin'                          /*define an array element for the coin.*/
          if _=='½' | _=="1/2"   then _= .5      /*an alternate spelling of a half─cent.*/
          if _=='¼' | _=="1/4"   then _= .25     /* "     "         "     " " quarter─¢.*/
          if \isNum(_)  then call ser ? "coin value isn't numeric."
          if _<0        then call ser ? "coin value can't be negative."
          if _<=0       then call ser ? "coin value can't be zero."
          if @._        then call ser ? "coin was already specified."
          if _<p        then call ser ? "coin must be greater than previous:"    p
          if _>N        then call ser ? "coin must be less or equal to amount:"  X
          @._= 1;   p= _                         /*signify coin was specified; set prev.*/
          $.j= _                                 /*assign the value to a particular coin*/
          end   /*j*/
_= n // 100;                    cnt= ' cents'    /* [↓]  is the amount in whole dollars?*/
if _=0  then do;   NN= '$'  ||  (NN%100);  cnt=  /*show the amount in dollars, not cents*/
             end
say 'with an amount of '        comma(NN)cnt",  there are "       comma( MKchg(N, coins) )
say 'ways to make change with coins of the following denominations: '    $
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isNum: return datatype(arg(1), 'N')              /*return 1 if arg is numeric, 0 if not.*/
ser:   say;   say '***error***';   say;   say arg(1);   say;   exit 13      /*error msg.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
comma: procedure; parse arg _;     n= _'.9';      #= 123456789;       b= verify(n, #, "M")
       e= verify(n, #'0', , verify(n, #"0.", 'M'))  -  4
             do j=e  to b  by -3;   _= insert(',', _, j);    end  /*j*/;          return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
MKchg: procedure expose $. !.;           parse arg a,k       /*function is recursive.   */
       if !.a.k\==.  then return !.a.k                       /*found this A & K before? */
       if a==0       then return 1                           /*unroll for a special case*/
       if k==1       then return 1                           /*   "    "  "    "      " */
       if k==2  then f= 1                                    /*handle this special case.*/
                else f= MKchg(a, k-1)                        /*count, recurse the amount*/
       if a==$.k  then do;  !.a.k= f+1;  return !.a.k;  end  /*handle this special case.*/
       if a <$.k  then do;  !.a.k= f  ;  return f    ;  end  /*   "     "     "      "  */
       !.a.k= f + MKchg(a-$.k, k);       return !.a.k        /*compute, define, return. */
output   is the same as the previous REXX version.



Ring

penny = 1
nickel = 1
dime = 1 
quarter = 1
count = 0
 
for penny = 0 to 100
    for nickel = 0 to 20
        for dime = 0 to 10
            for quarter = 0 to 4
                if (penny + nickel * 5 + dime * 10 + quarter * 25) = 100
                   see "" + penny + " pennies " + nickel + " nickels " + dime + " dimes " + quarter + " quarters" + nl
                   count = count + 1 
                ok
            next
        next
    next
next
see  count + " ways to make a dollar" + nl

Output:

0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
0 pennies 1 nickels 2 dimes 3 quarters
......
65 pennies 5 nickels 1 dimes 0 quarters
65 pennies 7 nickels 0 dimes 0 quarters
70 pennies 0 nickels 3 dimes 0 quarters
70 pennies 1 nickels 0 dimes 1 quarters

242 ways to make a dollar

RPL

Dynamic programming (space optimized)

Source: GeeksforGeeks website

« → coins sum
  « sum 1 + 1 →LIST 0 CON   @ dp[ii] will be storing the # of solutions for ii-1
    1 1 PUT                 @ base case
    1 coins SIZE FOR ii
       coins ii GET SWAP
       IF OVER sum ≤ THEN
          @ Pick all coins one by one and update dp[] values 
          @ after the index greater than or equal to the value of the picked coin 
          OVER 1 + sum 1 + FOR j
              DUP j GET
              OVER j 5 PICK - GET +
              j SWAP PUT
          NEXT 
       END SWAP DROP
    NEXT 
    DUP SIZE GET            
» » 'COUNT' STO
{ 1 5 10 25 } 100 COUNT
Output:
1: 242

Ruby

The algorithm also appears here

Recursive, with caching

def make_change(amount, coins)
  @cache = Array.new(amount+1){|i| Array.new(coins.size, i.zero? ? 1 : nil)}
  @coins = coins
  do_count(amount, @coins.length - 1)
end

def do_count(n, m)
  if n < 0 || m < 0
    0
  elsif @cache[n][m]
    @cache[n][m]
  else
    @cache[n][m] = do_count(n-@coins[m], m) + do_count(n, m-1)
  end
end

p make_change(   1_00, [1,5,10,25])
p make_change(1000_00, [1,5,10,25,50,100])

outputs

242
13398445413854501

Iterative

def make_change2(amount, coins)
  n, m = amount, coins.size
  table = Array.new(n+1){|i| Array.new(m, i.zero? ? 1 : nil)}
  for i in 1..n
    for j in 0...m
      table[i][j] = (i<coins[j] ? 0 : table[i-coins[j]][j]) +
                    (j<1        ? 0 : table[i][j-1])
    end
  end
  table[-1][-1]
end

p make_change2(   1_00, [1,5,10,25])
p make_change2(1000_00, [1,5,10,25,50,100])

outputs

242
13398445413854501

Run BASIC

for penny         = 0 to 100
  for nickel      = 0 to 20
    for dime      = 0 to 10
      for quarter = 0 to 4
       if penny + nickel * 5 + dime * 10 + quarter * 25 = 100 then
        print penny;" pennies ";nickel;" nickels "; dime;" dimes ";quarter;" quarters"
        count = count + 1 
      end if
      next quarter
    next dime
  next nickel
next penny
print count;" ways to make a buck"

Output:

0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
0 pennies 1 nickels 2 dimes 3 quarters
......
65 pennies 5 nickels 1 dimes 0 quarters
65 pennies 7 nickels 0 dimes 0 quarters
70 pennies 0 nickels 3 dimes 0 quarters
70 pennies 1 nickels 0 dimes 1 quarters
.....
242 ways to make a buck

Rust

fn make_change(coins: &[usize], cents: usize) -> usize {
    let size = cents + 1;
    let mut ways = vec![0; size];
    ways[0] = 1;
    for &coin in coins {
        for amount in coin..size {
            ways[amount] += ways[amount - coin];
        }
    }
    ways[cents]
}

fn main() {
    println!("{}", make_change(&[1,5,10,25], 100));
    println!("{}", make_change(&[1,5,10,25,50,100], 100_000));
}
Output:
242
13398445413854501

SAS

Generate the solutions using CLP solver in SAS/OR:

/* call OPTMODEL procedure in SAS/OR */
proc optmodel;
   /* declare set and names of coins */
   set COINS = {1,5,10,25};
   str name {COINS} = ['penny','nickel','dime','quarter'];

   /* declare variables and constraint */
   var NumCoins {COINS} >= 0 integer;
   con Dollar:
      sum {i in COINS} i * NumCoins[i] = 100;

   /* call CLP solver */
   solve with CLP / findallsolns;

   /* write solutions to SAS data set */
   create data sols(drop=s) from [s]=(1.._NSOL_) {i in COINS} <col(name[i])=NumCoins[i].sol[s]>;
quit;

/* print all solutions */
proc print data=sols;
run;

Output:

Obs penny nickel dime quarter 
1 100 0 0 0 
2 95 1 0 0 
3 90 2 0 0 
4 85 3 0 0 
5 80 4 0 0 
...
238 5 2 1 3 
239 0 3 1 3 
240 5 0 2 3 
241 0 1 2 3 
242 0 0 0 4 

Scala

def countChange(amount: Int, coins:List[Int]) = {
	  val ways = Array.fill(amount + 1)(0)
	  ways(0) = 1
	  coins.foreach (coin =>
	  for (j<-coin to amount)
		  ways(j) =  ways(j) + ways(j - coin)
		  )
	ways(amount)
  }       

countChange (15, List(1, 5, 10, 25))

Output:

res0: Int = 6

Recursive implementation:

def count(target: Int, coins: List[Int]): Int = {
  if (target == 0) 1
  else if (coins.isEmpty || target < 0) 0
  else count(target, coins.tail) + count(target - coins.head, coins)
}


count(100, List(25, 10, 5, 1))

Scheme

A simple recursive implementation:

(define ways-to-make-change
  (lambda (x coins)
    (cond
      [(null? coins) 0]
      [(< x 0) 0]
      [(zero? x) 1]
      [else (+ (ways-to-make-change x (cdr coins)) (ways-to-make-change (- x (car coins)) coins))])))

(ways-to-make-change 100)

Output:

242

Scilab

Straightforward solution

Fairly simple solution for the task. Expanding it to the optional task is not recommend, for Scilab will spend a lot of time processing the nested for loops.

amount=100;
coins=[25 10 5 1];
n_coins=zeros(coins);
ways=0;

for a=0:4
    for b=0:10
        for c=0:20
            for d=0:100
                n_coins=[a b c d];
                change=sum(n_coins.*coins);
                if change==amount then
                    ways=ways+1;
                elseif change>amount
                    break
                end
            end
        end
    end
end

disp(ways);
Output:
   242.

Faster approach

Translation of: Python
function varargout=changes(amount, coins)
    ways = zeros(1,amount + 2);
    ways(1) = 1;
    for coin=coins
        for j=coin:(amount+1)
            ways(j+1) = ways(j+1) + ways(j + 1 - coin);
        end
    end
    
    varargout=list(ways(length(ways)))
endfunction

a=changes(100, [1, 5, 10, 25]);
b=changes(100000, [1, 5, 10, 25, 50, 100]);
mprintf("%.0f, %.0f", a, b);
Output:
242, 13398445413854540

Seed7

$ include "seed7_05.s7i";
  include "bigint.s7i";
 
const func bigInteger: changeCount (in integer: amountCents, in array integer: coins) is func
  result
    var bigInteger: waysToChange is 0_;
  local
    var array bigInteger: t is 0 times 0_;
    var integer: pos is 0;
    var integer: s is 0;
    var integer: i is 0;
  begin
    t := length(coins) times 1_ & (length(coins) * amountCents) times 0_;
    pos := length(coins) + 1;
    for s range 1 to amountCents do
      if coins[1] <= s then
        t[pos] := t[pos - (length(coins) * coins[1])];
      end if;
      incr(pos);
      for i range 2 to length(coins) do
        if coins[i] <= s then
          t[pos] := t[pos - (length(coins) * coins[i])];
        end if;
        t[pos] +:= t[pos - 1];
        incr(pos);
      end for;
    end for;
    waysToChange := t[pos - 1];
  end func;
 
const proc: main is func
  local
    const array integer: usCoins is [] (1, 5, 10, 25, 50, 100);
    const array integer: euCoins is [] (1, 2, 5, 10, 20, 50, 100, 200);
  begin
    writeln(changeCount(    100, usCoins[.. 4]));
    writeln(changeCount( 100000, usCoins));
    writeln(changeCount(1000000, usCoins));
    writeln(changeCount( 100000, euCoins));
    writeln(changeCount(1000000, euCoins));
  end func;

Output:

242
13398445413854501
1333983445341383545001
10056050940818192726001
99341140660285639188927260001

SETL

program count_the_coins;
    print(count([1, 5, 10, 25], 100));
    print(count([1, 5, 10, 25, 50, 100], 1000 * 100));

    proc count(coins, n);
        tab := {[0, 1]};
        loop for coin in coins do
            loop for i in [coin..n] do
                tab(i) +:= tab(i - coin) ? 0;
            end loop;
        end loop;
        return tab(n);
    end proc;
end program;
Output:
242
13398445413854501

Sidef

Translation of: Perl
func cc(_)                { 0 }
func cc({ .is_neg  }, *_) { 0 }
func cc({ .is_zero }, *_) { 1 }

func cc(amount, first, *rest) is cached {
    cc(amount, rest...) + cc(amount - first, first, rest...);
}

func cc_optimized(amount, *rest) {
    cc(amount, rest.sort_by{|v| -v }...);
}

var x = cc_optimized(100, 1, 5, 10, 25);
say "Ways to change $1 with common coins: #{x}";

var y = cc_optimized(1000 * 100, 1, 5, 10, 25, 50, 100);
say "Ways to change $1000 with addition of less common coins: #{y}";
Output:
Ways to change $1 with common coins: 242
Ways to change $1000 with addition of less common coins: 13398445413854501

Swift

Translation of: Python
import BigInt

func countCoins(amountCents cents: Int, coins: [Int]) -> BigInt {
  let cycle = coins.filter({ $0 <= cents }).map({ $0 + 1 }).max()! * coins.count
  var table = [BigInt](repeating: 0, count: cycle)

  for x in 0..<coins.count {
    table[x] = 1
  }

  var pos = coins.count

  for s in 1..<cents+1 {
    for i in 0..<coins.count {
      if i == 0 && pos >= cycle {
        pos = 0
      }

      if coins[i] <= s {
        let q = pos - coins[i] * coins.count
        table[pos] = q >= 0 ? table[q] : table[q + cycle]
      }

      if i != 0 {
        table[pos] += table[pos - 1]
      }

      pos += 1
    }
  }

  return table[pos - 1]
}

let usCoins = [100, 50, 25, 10, 5, 1]
let euCoins = [200, 100, 50, 20, 10, 5, 2, 1]

for set in [usCoins, euCoins] {
  print(countCoins(amountCents: 100, coins: Array(set.dropFirst(2))))
  print(countCoins(amountCents: 100000, coins: set))
  print(countCoins(amountCents: 1000000, coins: set))
  print(countCoins(amountCents: 10000000, coins: set))
  print()
}
Output:
242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

Tailspin

Translation of: Rust
templates makeChange&{coins:}
  def paid: $;
  @: [1..$paid -> 0];
  $coins... -> \(def coin: $;
    @makeChange($coin): $@makeChange($coin) + 1;
    $coin+1..$paid -> @makeChange($): $@makeChange($) + $@makeChange($-$coin);
  \) -> !VOID
  $@($paid)!
end makeChange

100 -> makeChange&{coins: [1,5,10,25]} -> '$; ways to change a dollar
' -> !OUT::write
100000 -> makeChange&{coins: [1,5,10,25,50,100]} -> '$; ways to change 1000 dollars with all coins
' -> !OUT::write
Output:
242 ways to change a dollar
13398445413854501 ways to change 1000 dollars with all coins

Tcl

Translation of: Ruby
package require Tcl 8.5

proc makeChange {amount coins} {
    set table [lrepeat [expr {$amount+1}] [lrepeat [llength $coins] {}]]
    lset table 0 [lrepeat [llength $coins] 1]
    for {set i 1} {$i <= $amount} {incr i} {
	for {set j 0} {$j < [llength $coins]} {incr j} {
	    set k [expr {$i - [lindex $coins $j]}]
	    lset table $i $j [expr {
		($k < 0 ? 0 : [lindex $table $k $j]) +
		($j < 1 ? 0 : [lindex $table $i [expr {$j-1}]])
	    }]
	}
    }
    return [lindex $table end end]
}

puts [makeChange 100 {1 5 10 25}]
puts [makeChange 100000 {1 5 10 25 50 100}]
# Making change with the EU coin set:
puts [makeChange 100 {1 2 5 10 20 50 100 200}]
puts [makeChange 100000 {1 2 5 10 20 50 100 200}]

Output:

242
13398445413854501
4563
10056050940818192726001

uBasic/4tH

Translation of: Run BASIC
c = 0
for p       = 0 to 100
  for n     = 0 to 20
    for d   = 0 to 10
      for q = 0 to 4
       if p + n * 5 + d * 10 + q * 25 = 100 then
         print p;" pennies ";n;" nickels "; d;" dimes ";q;" quarters"
         c = c + 1
       endif
      next q
    next d
  next n
next p
print c;" ways to make a buck"
Output:
0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
...
90 pennies 2 nickels 0 dimes 0 quarters
95 pennies 1 nickels 0 dimes 0 quarters
100 pennies 0 nickels 0 dimes 0 quarters
242 ways to make a buck

0 OK, 0:312

UNIX Shell

Translation of: Common Lisp
Works with: bash
function count_change {
  local -i amount=$1 coin j
  local ways=(1)
  shift
  for coin; do
    for (( j=coin; j <= amount; j++ )); do
      let ways[j]=${ways[j]:-0}+${ways[j-coin]:-0}
    done
  done
  echo "${ways[amount]}"
}
count_change 100 25 10 5 1
count_change 100000 100 50 25 10 5 1
Works with: ksh version 93
function count_change {
  typeset -i amount=$1 coin j
  typeset ways
  set -A ways 1
  shift
  for coin; do
    for (( j=coin; j <= amount; j++ )); do
      let ways[j]=${ways[j]:-0}+${ways[j-coin]:-0}
    done
  done
  echo "${ways[amount]}"
}
count_change 100 25 10 5 1
count_change 100000 100 50 25 10 5 1
Works with: ksh version 88
function count_change {
  typeset -i amount=$1 coin j
  typeset ways
  set -A ways 1
  shift
  for coin; do
    let j=coin
    while (( j <= amount )); do
      let ways[j]=${ways[j]:-0}+${ways[j-coin]:-0}
      let j+=1
    done
  done
  echo "${ways[amount]}"
}
count_change 100 25 10 5 1
# (optional task exceeds a subscript limit in ksh88)

And just for fun, here's one that works even with the original V7 shell:

Works with: sh version v7
if [ $# -lt 2 ]; then
  set ${1-100} 25 10 5 1
fi
amount=$1
shift
ways_0=1
for coin in "$@"; do
  j=$coin
  while [ $j -le $amount ]; do
    d=`expr $j - $coin`
    eval "ways_$j=\`expr \${ways_$j-0} + \${ways_$d-0}\`"
    j=`expr $j + 1`
  done
done
eval "echo \$ways_$amount"
Output:
242
13398445413854501

VBA

Translation of: Phix
Private Function coin_count(coins As Variant, amount As Long) As Variant 'return type will be Decimal
    'sequence s = Repeat(0, amount + 1)
    Dim s As Variant
    ReDim s(amount + 1)
    Dim c As Integer
    s(1) = CDec(1)
    For c = 1 To UBound(coins)
        For n = coins(c) To amount
            s(n + 1) = CDec(s(n + 1) + s(n - coins(c) + 1))
        Next n
    Next c
    coin_count = s(amount + 1)
End Function
Public Sub main2()
    Dim us_commons_coins As Variant
    'The next line creates a base 1 array
    us_common_coins = [{25, 10, 5, 1}]
    Debug.Print coin_count(us_common_coins, 100)
    Dim us_coins As Variant
    us_coins = [{100,50,25, 10, 5, 1}]
    Debug.Print coin_count(us_coins, 100000)
End Sub
Output:
 242 
 13398445413854501 

VBScript

Translation of: C#
Function count(coins,m,n)
	ReDim table(n+1)
	table(0) = 1
	i = 0
	Do While i < m
		j = coins(i)
		Do While j <= n
			table(j) = table(j) + table(j - coins(i))
			j = j + 1
		Loop
		i = i + 1
	Loop
	count = table(n)
End Function

'testing
arr = Array(1,5,10,25)
m = UBound(arr) + 1
n = 100
WScript.StdOut.WriteLine count(arr,m,n)
Output:
242

Visual Basic

Translation of: VBA
Works with: Visual Basic version 6
Option Explicit
'----------------------------------------------------------------------
Private Function coin_count(coins As Variant, amount As Long) As Variant
'return type will be Decimal
Dim s() As Variant
Dim n As Long, c As Long
    
  ReDim s(amount + 1)
  s(1) = CDec(1)
  For c = LBound(coins) To UBound(coins)
    For n = coins(c) To amount
      s(n + 1) = CDec(s(n + 1) + s(n - coins(c) + 1))
    Next n
  Next c
  coin_count = s(amount + 1)
End Function
'----------------------------------------------------------------------
Sub Main()
Dim us_common_coins As Variant
Dim us_coins As Variant
    
  'The next line creates 0-based array
  us_common_coins = Array(25, 10, 5, 1)
  Debug.Print coin_count(us_common_coins, 100)
  
  us_coins = Array(100, 50, 25, 10, 5, 1)
  Debug.Print coin_count(us_coins, 100000)
  
End Sub
Output:
 242 
 13398445413854501

V (Vlang)

Translation of: Go
fn main() {
    amount := 100
    println("amount: $amount; ways to make change: ${count_change(amount)}")
}
 
fn count_change(amount int) i64 {
	if amount.str().count('0') > 4 {exit(-1)} // can be too slow
    return cc(amount, 4)
}
 
fn cc(amount int, kinds_of_coins int) i64 {
    if amount == 0 {return 1} 
    else if amount < 0 || kinds_of_coins == 0 {return 0}
    return cc(amount, kinds_of_coins-1) +
        cc(amount - first_denomination(kinds_of_coins), kinds_of_coins)
}
 
fn first_denomination(kinds_of_coins int) int {
    match kinds_of_coins {
		1 {return 1}
		2 {return 5}
		3 {return 10}
		4 {return 25}
		else {exit(-2)} 
	}
	return kinds_of_coins
}

Output:

amount, ways to make change: 100 242

Alternate:

fn main() {
	amount := 100
	coins := [25, 10, 5, 1]
	println("amount: $amount; ways to make change: ${count(coins, amount)}")
}

fn count(coins []int, amount int) int {
	mut ways := []int{len: amount + 1}
	ways[0] = 1
	for coin in coins {
		for idx := coin; idx <= amount; idx++ {
			ways[idx] += ways[idx - coin]
		}
	}
	return ways[amount]
}

Output:

amount: 100; ways to make change: 242

Wren

Translation of: Kotlin
Library: Wren-big
Library: Wren-fmt
import "./big" for BigInt
import "./fmt" for Fmt

var countCoins = Fn.new { |c, m, n|
    var table = List.filled(n + 1, null)
    table[0] = BigInt.one
    for (i in 1..n) table[i] = BigInt.zero
    for (i in 0...m) {
        for (j in c[i]..n) table[j] = table[j] + table[j-c[i]]
    }
    return table[n]
}

var c = [1, 5, 10, 25, 50, 100]
Fmt.print("Ways to make change for $$1 using 4 coins     = $,i", countCoins.call(c, 4, 100))
Fmt.print("Ways to make change for $$1,000 using 6 coins = $,i", countCoins.call(c, 6, 1000 * 100))
Output:
Ways to make change for $1 using 4 coins     = 242
Ways to make change for $1,000 using 6 coins = 13,398,445,413,854,501

zkl

Translation of: Scheme
fcn ways_to_make_change(x, coins=T(25,10,5,1)){
   if(not coins) return(0);
   if(x<0)  return(0);
   if(x==0) return(1);
   ways_to_make_change(x, coins[1,*]) + ways_to_make_change(x - coins[0], coins)
}
ways_to_make_change(100).println();
Output:
242

Blows the stack on the optional part, so try this:

Translation of: Ruby
fcn make_change2(amount, coins){
  n, m  := amount, coins.len();
  table := (0).pump(n+1,List, (0).pump(m,List().write,1).copy);
  foreach i,j in ([1..n],[0..m-1]){
     table[i][j] = (if(i<coins[j]) 0 else table[i-coins[j]][j]) +
                   (if(j<1)        0 else table[i][j-1])
  }
  table[-1][-1]
}

println(make_change2(   100, T(1,5,10,25)));
make_change2(0d1000_00, T(1,5,10,25,50,100)) : "%,d".fmt(_).println();
Output:
242
13,398,445,413,854,501

ZX Spectrum Basic

Translation of: AWK

Test with emulator at full speed for reasonable performance.

10 LET amount=100
20 GO SUB 1000
30 STOP 
1000 LET nPennies=amount
1010 LET nNickles=INT (amount/5)
1020 LET nDimes=INT (amount/10)
1030 LET nQuarters=INT (amount/25)
1040 LET count=0
1050 FOR p=0 TO nPennies
1060 FOR n=0 TO nNickles
1070 FOR d=0 TO nDimes
1080 FOR q=0 TO nQuarters
1090 LET s=p+n*5+d*10+q*25
1100 IF s=100 THEN LET count=count+1
1110 NEXT q
1120 NEXT d
1130 NEXT n
1140 NEXT p
1150 PRINT count
1160 RETURN