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There are four types of common coins in US currency:

quarters (25 cents)
dimes (10 cents)
nickels (5 cents), and
pennies (1 cent)

There are six ways to make change for 15 cents:

A dime and a nickel
A dime and 5 pennies
3 nickels
2 nickels and 5 pennies
A nickel and 10 pennies
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

an algorithm from the book Structure and Interpretation of Computer Programs.
an article in the algorithmist.
Change-making problem on Wikipedia.

11l

Translation of: Python

<lang 11l>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]))</lang>

Output:

242
13398445413854501

Ada

Works with: gnat/gcc

<lang Ada>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;</lang>

Output:

 242
 13398445413854501

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 </lang>

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 </lang>

Output:

                                +242
                       +139946140451
                  +13398445413854501

AppleScript

Translation of: Phix

<lang applescript>-- 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}</lang>

Output:

Arturo

Translation of: Python

<lang arturo>changes: @(amount coins){ ways: #(0)*(amount+1) ways.0: 1

loop coins @(coin){ loop coin..amount @(j){ ways.[j]: ways.[j]+ways.[j-coin] } } return ways.[amount] }

print [changes 100 #(1 5 10 25)] print [changes 100000 #(1 5 10 25 50 100)]</lang>

Output:

242
13398445413854501

AutoHotkey

Translation of: Go

Works with: AutoHotkey_L

<lang AHK>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)</lang>

AWK

Iterative implementation, derived from Run BASIC:

<lang awk>#!/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;

} </lang>

Run time:

time ./change-itr.awk
242

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

Recursive implementation (derived from Scheme example):

<lang awk>#!/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]

} </lang>

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: <lang bbcbasic> 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)

</lang> 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. <lang c>#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) Template: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; }</lang>output (only the first two lines are required by task):<lang>242 13398445413854501 1333983445341383545001 133339833445334138335450001

4563 10056050940818192726001 99341140660285639188927260001 992198221207406412424859964272600001</lang>

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();
       }
   }

</lang>

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;

}</lang>

Output:

Clojure

<lang lisp>(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 </lang>

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
          .

</lang>

Output:

Coco

Translation of: Python

<lang coco>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]</lang>

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:

Reversed the order of the loops to start counting with the largest denomination > smallest denomination. Result: 44:45
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.
Not printing any of the individual results speeds up total time to 9:30.
Removing the specific variables used in the NEXT statements helps the interpreter speed up. Result: 9:10.
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.

<lang gwbasic>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</lang>

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.

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:

Common Lisp

Recursive Version With Cache

<lang lisp>(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)</lang>

Iterative Version

<lang lisp>(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))</lang>

D

Basic Version

Translation of: Go

<lang d>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;

}</lang>

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. <lang d>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;

}</lang> The output is the same.

Faster Version

Translation of: C

<lang d>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;
   }

}</lang>

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

<lang d>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;
   }

}</lang>

Printing Version

This version prints all the solutions (so it can be used on the smaller input): <lang d>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]);

}</lang>

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. <lang Dart> 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);
   }

</lang>

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

Delphi

Translation of: C#

<lang Delphi> 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. </lang>

Dyalect

<lang 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))</lang>

Output:

EchoLisp

Recursive solution using memoization, adapted from CommonLisp and Racket. <lang scheme> (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 </lang>

Output:

<lang scheme> (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

</lang>

Elixir

Recursive Dynamic Programming solution in Elixir <lang 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)</lang>

Output:

242
13398445413854501

Erlang

<lang 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).

</lang>

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.

<lang fsharp>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</lang>

Output:

242
13398445413854501

Factor

<lang 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 ;</lang>

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: <lang> IN: scratchpad [ 100000 { 1 5 10 25 50 100 } make-change . ] time 13398445413854501 Running time: 0.020869274 seconds </lang> For reference, the implementation is shown next. <lang> 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) ;

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

Forth

<lang 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 .</lang>

FreeBASIC

Translation from "Dynamic Programming Solution: Python version" on this webside [1] <lang freebasic>' 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</lang>

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

<lang futurebasic> include "ConsoleWindow"

dim as long penny, nickel, dime, quarter , count

penny = 1 : nickel = 1 dime = 1 : quarter = 1

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
              print penny; " pennies "; nickel;" nickels "; dime; " dimes "; quarter; " quarters"
              count++ 
           end if
        next quarter
     next dime
  next nickel

next penny print count;" ways to make a dollar"

</lang>

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

A translation of the Lisp code referenced by the task description: <lang go>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)

}</lang> Output:

amount, ways to make change: 100 242

Alternative algorithm, practical for the optional task. <lang go>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]

}</lang> Output:

amount, ways to make change: 100000 13398445413854501

Groovy

Translation of: Go

Intuitive Recursive Solution: <lang groovy>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)
   }

}</lang>

Fast Iterative Solution: <lang groovy>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]

}</lang>

Test: <lang groovy>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")</lang>

Output:

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

Extra Credit:
iterative answer: 13398445413854501   elapsed: 1077ms

Haskell

Naive implementation: <lang haskell>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])</lang>

Much faster, probably harder to read, is to update results from bottom up: <lang haskell>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)</lang>

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

<lang haskell>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)
   ]

</lang>

Output:

242
139946140451
1333983445341383545001

Icon and Unicon

<lang Icon>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</lang>

This is a naive implementation and very slow.

This example is in need of improvement:

Needs a better algorithm.

<lang Icon>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</lang>

Library: Icon Programming Library

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: <lang Icon>

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 </lang> 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

<lang 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."</lang>

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).

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

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:

<lang j> 100 nsplits 1 5 10 25 242</lang>

And, on a 64 bit machine with sufficient memory:

<lang j> 100000 nsplits 1 5 10 25 50 100 13398445413854501</lang>

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

<lang j> 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)</lang>

Java

Translation of: D

Works with: Java version 1.5+

<lang java5>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");
       }
   }

}</lang> Output:

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

JavaScript

Iterative

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

<lang Javascript>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];

}</lang>

Output:

JavaScript hits integer limit for optional task <lang JavaScript>countcoins(100, [1,5,10,25]); 242</lang>

Recursive

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

<lang Javascript>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;

}</lang>

Output:

Too slow for optional task

<lang JavaScript>countcoins(100, [1,5,10,25]); 242</lang>

Iterative again

Translation of: C#

<lang javascript>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])</lang>

Output:

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. <lang jq># 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] ;</lang>

Example:

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

Output:

Julia

Translation of: Python

<lang julia>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])</lang>

Output:

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

Kotlin

Translation of: C#

<lang scala>// 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))

}</lang>

Output:

242
13398445413854501

Lasso

Inspired by the javascript iterative example for the same task <lang Lasso>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,)) '
' cointcoins(100000, array(1, 5, 10, 25, 50, 100))</lang> 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"... <lang Lua>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}))</lang>

Output:

242
1.3398445413855e+16

M2000 Interpreter

Fast O(n*m)

Works with decimals in table() <lang M2000 Interpreter> 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 </lang>

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)

<lang M2000 Interpreter> 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 </lang>

Maple

Straightforward implementation with power series. Not very efficient for large amounts.

<lang maple>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</lang>

A faster implementation.

<lang maple>ways2:=proc(amount,purse)

 local a,n,k;
 a:=Array(1..amount,0);
 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(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</lang>

Mathematica / Wolfram Language

Translation of: Go

<lang Mathematica>CountCoins[amount_, coinlist_] := ( ways = ConstantArray[1, amount]; Do[For[j = coin, j <= amount, j++,

 If[ j - coin == 0,
   waysj ++,
   waysj += waysj - coin

]] , {coin, coinlist}]; waysamount)</lang> Example usage:

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

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

MATLAB / Octave

<lang Matlab> %% 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 </lang> Example Output:

Main Challenge: 242

Bonus Challenge: 13398445413854501

Mercury

<lang 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).</lang>

Nim

Translation of: Python

<lang nim>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])</lang> Output:

242
13398445413854501

OCaml

Translation of the D minimal version:

<lang ocaml>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]);

</lang>

Output:

$ ocaml coins.ml 
242
13398445413854501

PARI/GP

<lang parigp>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)</lang> Output:

%1 = 242
%2 = 13398445413854501

Perl

<lang 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 );

</lang>

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 <lang Phix>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</lang> An attempt to explain this algorithm further seems worthwhile: <lang Phix>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))</lang>

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

PicoLisp

Translation of: C

<lang PicoLisp>(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 ) )</lang>

Test: <lang PicoLisp>(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) )</lang>

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.

<lang Prolog>:- 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]).</lang>

Output:

?- aggregate(count, coins_for(100), Count).
Count = 242.

Python

Simple version

Translation of: Go

<lang python>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])</lang> Output:

242
13398445413854501

Fast version

Translation of: C

<lang python>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()</lang> Output:

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

Racket

This is the basic recursive way: <lang Racket>#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 </lang> 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>#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 |#</lang>

Raku

(formerly Perl 6)

Works with: rakudo version 2018.10

Translation of: Ruby

<lang perl6># 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];</lang>

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). <lang rexx>/*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.*/</lang>

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 1^st version when using larger amounts. <lang rexx>/*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. */</lang>

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. <lang rexx>/*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 _

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. */</lang>

output is the same as the previous REXX version.

Ring

<lang 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 </lang> 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

Ruby

The algorithm also appears here

Recursive, with caching

<lang ruby>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])</lang>

outputs

242
13398445413854501

Iterative

<lang ruby>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])</lang> outputs

242
13398445413854501

Run BASIC

<lang runbasic>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"</lang>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

<lang 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));

}</lang>

Output:

242
13398445413854501

SAS

Generate the solutions using CLP solver in SAS/OR: <lang sas>/* 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;</lang>

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

<lang 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)) </lang> Output:

res0: Int = 6

Recursive implementation: <lang scala>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)) </lang>

Scheme

A simple recursive implementation: <lang scheme>(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)</lang> Output:

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. <lang>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);</lang>

Output:

   242.

Faster approach

Translation of: Python

<lang>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);</lang>

Output:

242, 13398445413854540

Seed7

<lang 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;</lang>

Output:

242
13398445413854501
1333983445341383545001
10056050940818192726001
99341140660285639188927260001

Sidef

Translation of: Perl

<lang ruby>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}";</lang>

Output:

Ways to change $1 with common coins: 242
Ways to change $1000 with addition of less common coins: 13398445413854501

Swift

Translation of: Python

Library: Attaswift BigInt

<lang swift>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()

}</lang>

Output:

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

Tailspin

Translation of: Rust

<lang tailspin> 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 </lang>

Output:

242 ways to change a dollar
13398445413854501 ways to change 1000 dollars with all coins

Tcl

Translation of: Ruby

<lang tcl>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}]</lang> Output:

242
13398445413854501
4563
10056050940818192726001

uBasic/4tH

Translation of: Run BASIC

<lang>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"</lang>

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

<lang 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</lang>

Works with: ksh version 93

<lang bash>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</lang>

Works with: ksh version 88

<lang bash>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)</lang>

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

Works with: sh version v7

<lang bash>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"</lang>

Output:

242
13398445413854501

VBA

Translation of: Phix

<lang vb>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</lang>

Output:

 242 
 13398445413854501

VBScript

Translation of: C#

<lang vb> 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) </lang>

Output:

Visual Basic

Translation of: VBA

Works with: Visual Basic version 6

<lang vb>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</lang>

Output:

 242 
 13398445413854501

Wren

Translation of: Kotlin

Library: Wren-big

Library: Wren-fmt

<lang ecmascript>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))</lang>

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

<lang zkl>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();</lang>

Output:

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

Translation of: Ruby

<lang zkl>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();</lang>

Output:

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

ZX Spectrum Basic

Translation of: AWK

Test with emulator at full speed for reasonable performance. <lang zxbasic>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 </lang>