Count the coins
There are four types of common coins in US currency:
You are encouraged to solve this task according to the task description, using any language you may know.
- 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
<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
360 Assembly
<lang 360asm>* count the coins 04/09/2015 COINS CSECT
USING COINS,R12 LR R12,R15 L R8,AMOUNT npenny=amount L R4,AMOUNT SRDA R4,32 D R4,=F'5' LR R9,R5 nnickle=amount/5 L R4,AMOUNT SRDA R4,32 D R4,=F'10' LR R10,R5 ndime=amount/10 L R4,AMOUNT SRDA R4,32 D R4,=F'25' LR R11,R5 nquarter=amount/25 SR R1,R1 count=0 SR R4,R4 p=0
LOOPP CR R4,R8 do p=0 to npenny
BH ELOOPP SR R5,R5 n=0
LOOPN CR R5,R9 do n=0 to nnickle
BH ELOOPN SR R6,R6
LOOPD CR R6,R10 do d=0 to ndime
BH ELOOPD SR R7,R7 q=0
LOOPQ CR R7,R11 do q=0 to nquarter
BH ELOOPQ LR R3,R5 n MH R3,=H'5' LR R2,R4 p AR R2,R3 LR R3,R6 d MH R3,=H'10' AR R2,R3 LR R3,R7 q MH R3,=H'25' AR R2,R3 s=p+n*5+d*10+q*25 C R2,=F'100' if s=100 BNE NOTOK LA R1,1(R1) count=count+1
NOTOK LA R7,1(R7) q=q+1
B LOOPQ
ELOOPQ LA R6,1(R6) d=d+1
B LOOPD
ELOOPD LA R5,1(R5) n=n+1
B LOOPN
ELOOPN LA R4,1(R4) p=p+1
B LOOPP
ELOOPP XDECO R1,PG+0 edit count
XPRNT PG,12 print count XR R15,R15 BR R14
AMOUNT DC F'100' start value in cents PG DS CL12
YREGS END COINS</lang>
- Output:
242
Ada
<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
<lang Algol68>
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
<lang Algol68>
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
<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:
<lang applescript>{242, 13398445413854501}</lang>
Arturo
<lang rebol>changes: function [amount coins][ ways: map 0..amount+1 [x]-> 0 ways\0: 1
loop coins 'coin [ loop coin..amount 'j -> set ways j (get ways j) + get ways j-coin ]
ways\[amount] ]
print changes 100 [1 5 10 25] print changes 100000 [1 5 10 25 50 100]</lang>
AutoHotkey
<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#
<lang csharp>
// 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++
<lang cpp>
- 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:
242
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
<lang cobol>
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:
242
Coco
<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$,2);"."
270 end
1000 print count;tab(6);pc;tab(11);nc;tab(16);dc;tab(21);qc:return</lang>
Example 2: Commodore 64 with Screen Blanking
Make the following changes on a Commodore 64 to enable screen blanking. This will give the CPU a few extra cycles normally held by the VIC-II. Add line 145 and change line 245 as shown.
Enabling screen blanking (and therefore not printing each result) results in a total time of 1:44.
<lang gwbasic>145 if not yn then poke 53265,peek(53265) and 239 245 en$=ti$:if not yn then poke 53265,peek(53265) or 16</lang>
Example 3: Commodore 128 with VIC-II blanking, 2MHz fast mode.
Similar to above, however the Commodore 128 is capable of using a faster clock speed at the expense of any VIC-II graphics display. Timed result is 1:18. Add/change the following lines on the Commodore 128:
<lang gwbasic>145 if not yn then fast 245 en$=ti$:if not yn then slow</lang>
Common Lisp
Recursive Version With Cache
<lang lisp>(defun count-change (amount coins
&optional (length (1- (length coins))) (cache (make-array (list (1+ amount) (length coins)) :initial-element nil))) (cond ((< length 0) 0) ((< amount 0) 0) ((= amount 0) 1) (t (or (aref cache amount length) (setf (aref cache amount length) (+ (count-change (- amount (first coins)) coins length cache) (count-change amount (rest coins) (1- length) cache)))))))
- (compile 'count-change) ; for CLISP
(print (count-change 100 '(25 10 5 1))) ; = 242 (print (count-change 100000 '(100 50 25 10 5 1))) ; = 13398445413854501 (terpri)</lang>
Iterative Version
<lang lisp>(defun count-change (amount coins &aux (ways (make-array (1+ amount) :initial-element 0)))
(setf (aref ways 0) 1) (loop for coin in coins do (loop for j from coin upto amount do (incf (aref ways j) (aref ways (- j coin))))) (aref ways amount))</lang>
D
Basic Version
<lang d>import std.stdio, std.bigint;
auto changes(int amount, int[] coins) {
auto ways = new BigInt[amount + 1]; ways[0] = 1; foreach (coin; coins) foreach (j; coin .. amount + 1) ways[j] += ways[j - coin]; return ways[$ - 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
<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.
<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.$coinTypes"; // lookes like "100.[25,10,5,1]" var cacheValue = cache[key]; // check whether we have seen this before
if(cacheValue != null) return(cacheValue);
count = 0; // same recursion as simple method, but caches all subqueries too for(int i=0; i<=(amount/coinTypes[0]).toInt(); i++){ count += cachedCoins(amount-(i*coinTypes[0]),coinTypes.sublist(1)); // sublist(1) is like lisp's '(rest ...)' }
cache[key] = count; // add this to the cache return(count); }
</lang>
- Output:
242 ways for 100 using [25, 10, 5, 1] coins. 13398445413854501 ways for 100000 using [100, 50, 25, 10, 5, 1] coins. ... completed in 3.604 seconds
Delphi
<lang Delphi> program Count_the_coins;
{$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:
242
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>
EDSAC order code
The program solves the first task for the US dollar and UK pound, using an algorithm copied from the C# and Delphi solutions. The second task is not attempted.
Note: When the table is initialized, not only must the first entry be set to 1, but the other entries must be set to 0. It seems that the C# and Delphi solutions rely on the compiler to do this. In other languages, it may need to be done by the program. <lang edsac> ["Count the coins" problem for Rosetta Code.] [EDSAC program, Initial Orders 2.]
T51K P56F [G parameter: print subroutine] T54K P94F [C parameter: coins subroutine] T47K P200F [M parameter: main routine]
[========================== M parameter ===============================]
E25K TM GK
[Parameter block for US coins. For convenience, all numbers
are in the address field, e.g. 25 cents is P25F not P12D.] [0] UF SF [2-letter ID] P100F [amount to be made with coins] P4F [number of coin values] P1F P5F P10F P25F [list of coin values] [8] P@ [address of US parameter block]
[Parameter block for UK coins]
[9] UF KF P100F P7F P1F P2F P5F P10F P20F P50F P100F [20] P9@ [address of UK parameter block]
[Enter with acc = 0]
[21] A8@ [load address of parameter block for US coins] T4F [pass to subroutine in 4F] [23] A23@ [call subroutine to calculate and print result] G13C A20@ [same for UK coins] T4F [27] A27@ G13C ZF [halt program]
[========================== C parameter ===============================] [Subroutine to calculate and print the result for the given amount and
set of coins. Address of parameter block (see above) is passed in 4F.] E25K TC GK [0] SF [S order for start of coin list] [1] A1023F [start table at top of memory and work downwarda] [2] PF [S order for exclusive end of coin list] [3] P2F [to increment address by 2] [4] OF [(1) add to address to make O order (2) add to A order to make T order with same address] [5] SF [add to address to make S order] [6] K4095F [add to S order to make A order, dec address] [7] K2048F [set teleprinter to letters] [8] #F [set teleprinter to figures] [9] !F [space character] [10] @F [carriage return] [11] &F [line feed] [12] K4096F [teleprinter null]
[Subroutine entry. In this EDSAC program, the table used
in the algorithm grows downward from the top of memory.] [13] A3F [plant jump back to caller, as usual] T89@ A4F [load address of parameter block] A3@ [skip 2-letter ID] A5@ [make S order for amount] U27@ [plant in code] A3@ [make S order for first coin value] U@ [store it] A6@ [make A order for number of coins] T38@ [plant in code] A2F [load 1 (in address field)] [24] T1023F [store at start of table]
[Set all other table entries to 0]
A24@ T32@ [27] SF [acc := -amount] [28] TF [set negative count in 0F] A32@ [decrement address in manufactured order] S2F T32@ [32] TF [manufactured: set table entry to 0] AF [update negative count] A2F G28@ [loop until count = 0]
[Here acc = 0. Manufactured order (4 lines up) is T order
for inclusive end of table; this is used again below.] A@ [load S order for first coin value] U43@ [plant in code] [38] AF [make S order for exclusive end of coin list] T2@ [store for comparison]
[Start of outer loop, round coin values]
[40] TF [clear acc] A1@ [load A order for start of table] U48@ [plant in code] [43] SF [manufactured order: subtract coin value]
[Start of inner loop, round table entries]
[44] U47@ [plant A order in code] A4@ [make T order for same address] T49@ [plant in code]
[The next 3 orders are manufactured at run time]
[47] AF [load table entry] [48] AF [add earlier table entry] [49] TF [update table entry] A32@ [load T order for inclusive end of table] S49@ [reached end of table?] E60@ [if yes, jump out of inner loop] TF [clear acc] A48@ [update the 3 manufactured instructions] S2F T48@ A47@ S2F G44@ [always loops back, since A < 0]
[End of inner loop]
[60] TF [clear acc] A43@ [update S order for coin value] A2F U43@ S2@ [reached exclusive end?] G40@ [if no, loop back]
[End of outer loop] [Here with acc = 0 and result at end of table] [Value is in address field, so shift 1 right for printing]
A32@ [load T order for end of tab;e] S4@ [make A order for same address] T79@ [plant in code] A4F [load address of parameter block] A4@ [make O order for 1st char of ID] U75@ [plant in code] A2F [same for 2nd char] T76@ O7@ [set teleprinter to letters] [75] OF [print ID, followed by space] [76] OF O9@ O8@ [set teleprinter to figures] [79] AF [maunfactured order to load result] RD [shift 1 right for printing] TF [pass to print routine] A9@ [replace leading 0's with space] T1F [84] A84@ [call print routine] GG O10@ O11@ [print CR, LF] O12@ [print null to flush teleprinter buffer] [89] ZF [replaced by jump back to caller]
[============================= G parameter ===============================]
E25K TG GK
[Subroutine to print non-negative 17-bit integer. Always prints 5 chars.
Caller specifies character for leading 0 (typically 0, space or null). Parameters: 0F = integer to be printed (not preserved) 1F = character for leading zero (preserved) Workspace: 4F..7F, 38 locations] A3FT34@A1FT7FS35@T6FT4#FAFT4FH36@V4FRDA4#FR1024FH37@E23@O7FA2F T6FT5FV4#FYFL8FT4#FA5FL1024FUFA6FG16@OFTFT7FA6FG17@ZFP4FZ219DTF
[========================== M parameter again ===============================]
E25K TM GK E21Z [define entry point] PF [enter with acc = 0]
</lang>
- Output:
US 242 UK 4563
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#
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
<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
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.
<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>
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
<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
<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:
242
jq
Currently jq uses IEEE 754 64-bit numbers. Large integers are approximated by floats, and therefore the answer that the following program provides for the optional task is only correct for the first 15 digits. <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:
242
Julia
<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
<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. Note that in the following, all amounts are in cents.
<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); for k in purse do for n from k to amount do if n=k then a[n]++; else a[n]+=a[n-k] fi od od; a[-1]
end:
ways2(100,[1,5,10,25]);
- 242
ways2(100,[1,5,10,25,50,100]);
- 293
ways2(1000,[1,5,10,25,50,100]);
- 2103596
ways2(10000,[1,5,10,25,50,100]);
- 139946140451
ways2(100000,[1,5,10,25,50,100]);
- 13398445413854501
ways2(1000000,[1,5,10,25,50,100]);
- 1333983445341383545001
ways2(10000000,[1,5,10,25,50,100]);
- 133339833445334138335450001
ways2(100000000,[1,5,10,25,50,100]);
- 13333398333445333413833354500001</lang>
Additionally, while it's not proved as is, we can see that the first values for an amount 10^k obey the following simple formula:
<lang maple>P:=n->4/(3*10^9)*n^5+65/10^8*n^4+112/10^6*n^3+805/10^5*n^2+635/3000*n+1:
for k from 2 to 8 do lprint(P(10^k)) od: 293 2103596 139946140451 13398445413854501 1333983445341383545001 133339833445334138335450001 13333398333445333413833354500001</lang>
The polynomial P(n) seems to give the correct number of ways iff n is a multiple of 100 (tested up to n=10000000), i.e. the number of ways for 100n is
<lang maple>Q:=n->40/3*n^5+65*n^4+112*n^3+161/2*n^2+127/6*n+1:</lang>
Mathematica / Wolfram Language
<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
<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
function coin_count(sequence coins, integer amount) sequence s = repeat(0,amount+1) s[1] = 1 for c=1 to length(coins) do for n=coins[c] to amount do s[n+1] += s[n-coins[c]+1] end for end for return s[amount+1] end function
An attempt to explain this algorithm further seems worthwhile:
function coin_count(sequence coins, integer amount) -- start with 1 known way to achieve 0 (being no coins) -- (nb: s[1] holds the solution for 0, s[n+1] for n) sequence s = repeat(0,amount+1) s[1] = 1 -- then for every coin that we can use, increase number of -- solutions by that previously found for the remainder. for c=1 to length(coins) do -- this inner loop is essentially behaving as if we had -- called this routine with 1..amount, but skipping any -- less than the coin's value, hence coins[c]..amount. for n=coins[c] to amount do s[n+1] += s[n-coins[c]+1] end for end for return s[amount+1] end function -- The key to understanding the above is to try a dry run of this: printf(1,"%d\n",coin_count({2,3},5)) -- (prints 1) -- You'll need 4 2p coins, 3 3p coins, and 5 spaces marked 1..5. -- Place 2p wherever it fits: 1:0 2:1 3:1 4:1 5:1 -- Add previously found solns: +0 +1 +0 +1 +0 [1] -- Place 3p wherever it fits: 1:0 2:0 3:1 4:1 5:1 -- Add previously found solns: +0 +0 +1 +0 +1 [2] -- [1] obviously at 2: we added the base soln for amount=0, -- and at 4: we added the previously found soln for 2. -- also note that we added nothing for 2p+3p, yet, that -- fact is central to understanding why this works. [3] -- [2] obviously at 3: we added the base soln for amount=0, -- at 4: we added the zero solutions yet found for 1p, -- and at 5: we added the previously found soln for 2. -- you can imagine at 6,9,12 etc all add in soln for 3, -- albeit by adding that as just added to the precessor. -- [3] since we add no 3p solns when processing 2p, we do -- not count 2p+3p and 3p+2p as two solutions. --For N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. printf(1,"%d\n",coin_count({1,2,3},4)) --For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}. printf(1,"%d\n\n",coin_count({2,3,5,6},10)) printf(1,"%d\n",coin_count({25, 10, 5, 1},1_00)) printf(1,"%,d\n",coin_count({100, 50, 25, 10, 5, 1},1000_00))
- Output:
1 4 5 242 13,398,445,413,854,501
Note that a slightly wrong value is printed when running this on 32 bits:
13,398,445,413,854,501 -- 64 bit (exact) 13,398,445,413,854,496 -- 32 bit (5 out) 9,007,199,254,740,992 -- max precision (53 bits) of a 64-bit float
PicoLisp
<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
<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
<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
Quackery
<lang Quackery> [ stack ] is lim ( --> s )
[ swap dup 1+ lim put 1 0 rot of join swap witheach [ 0 over of swap negate temp put lim share times [ over i^ peek over temp share peek + join ] temp take negate split nip nip ] -1 peek lim release ] is makechange ( n [ --> n )
say "With US coins." cr 100 ' [ 1 5 10 25 ] makechange echo cr 100000 ' [ 1 5 10 25 50 100 ] makechange echo cr cr say "With EU coins." cr 100 ' [ 1 2 5 10 20 50 100 200 ] makechange echo cr 100000 ' [ 1 2 5 10 20 50 100 200 ] makechange echo cr</lang>
- Output:
With US coins. 242 13398445413854501 With EU coins. 4563 10056050940818192726001
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)
<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 1st 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:
242
Scilab
Straightforward solution
Fairly simple solution for the task. Expanding it to the optional task is not recommend, for Scilab will spend a lot of time processing the nested for
loops.
<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
<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
<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
<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
<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
<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
<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
<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>
<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>
<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:
<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
<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
<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:
242
Visual Basic
<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
Vlang
<lang go>fn main() {
amount := 100 println("amount, ways to make change: $amount ${count_change(amount)}"))
}
fn count_change(amount int) i64 {
return cc(amount, 4)
}
fn cc(amount int, kinds_of_coins int) i64 {
if amount == 0 { return 1 } else if amount < 0 || kinds_of_coins == 0 { return 0 } return cc(amount, kinds_of_coins-1) + cc(amount - first_denomination(kinds_of_coins), kinds_of_coins)
}
fn first_denomination(kinds_of_coins int) int {
match kinds_of_coins {
1 { return 1 } 2 { return 5 } 3 { return 10 } 4 { return 25 } } }</lang> Output:
amount, ways to make change: 100 242
Wren
<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
<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:
242
Blows the stack on the optional part, so try this:
<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
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>