如果给出一些美元价值，如何找到所有硬币组合

Question

几个月前，我找到了一段我正在编写的面试代码。

根据我的评论，它试图解决这个问题：

  

给定美分的美元价值（例如200 = 2美元，1000 = 10美元），找到构成美元价值的所有硬币组合。   只有便士（1¢），镍（5¢），角钱（10¢）和四分之一（25¢）。

例如，如果给出100，答案应该是：

4 quarter(s) 0 dime(s) 0 nickel(s) 0 pennies  
3 quarter(s) 1 dime(s) 0 nickel(s) 15 pennies  
etc.

我相信这可以通过迭代和递归方式解决。我的递归解决方案非常错误，我想知道其他人如何解决这个问题。这个问题的难点在于尽可能提高效率。

Answer 1

我很久以前就研究了这个问题，你可以阅读我的little write-up on it。这是Mathematica source。

通过使用生成函数，您可以获得一个封闭形式的常量时间解决方案。格雷厄姆，克努特和帕塔什尼克的具体数学就是这本书的书，并且对这个问题进行了相当广泛的讨论。基本上，您定义了一个多项式，其中 n th系数是对 n 美元进行更改的方式的数量。

该文章的第4-5页显示了如何使用Mathematica（或任何其他方便的计算机代数系统）在三行代码中在几秒钟内计算10 ^ 10 ^ 6美元的答案。

（而且这已经足够久了，75Mhz Pentium上只有几秒钟......）

Answer 2

注意：这仅显示方式的数量。

Scala函数：

def countChange(money: Int, coins: List[Int]): Int =
  if (money == 0) 1
  else if (coins.isEmpty || money < 0) 0
  else countChange(money - coins.head, coins) + countChange(money, coins.tail)

Answer 3

我赞成递归解决方案。你有一些面额列表，如果最小的面额可以平均分配剩余的货币金额，这应该可以正常工作。

基本上，你从最大面额移动到最小面额 递归，

1. 您有一个当前总数要填补，以及一个最大面额（剩下超过1个）。 如果只剩下1个面额，那么只有一种方法可以填补总数。您可以使用当前面额的0到k份副本，使k * cur面额<=总计。
2. 对于0到k，使用修改后的总计和新的最大面额调用函数。
3. 将结果从0添加到k。这就是你可以用多少种方式从目前的面额中填补总数。退回这个号码。

这是我所说的问题的python版本，200美分。我得到了1463种方式。此版本打印所有组合和最终计数总数。

#!/usr/bin/python

# find the number of ways to reach a total with the given number of combinations

cents = 200
denominations = [25, 10, 5, 1]
names = {25: "quarter(s)", 10: "dime(s)", 5 : "nickel(s)", 1 : "pennies"}

def count_combs(left, i, comb, add):
    if add: comb.append(add)
    if left == 0 or (i+1) == len(denominations):
        if (i+1) == len(denominations) and left > 0:
            comb.append( (left, denominations[i]) )
            i += 1
        while i < len(denominations):
            comb.append( (0, denominations[i]) )
            i += 1
        print(" ".join("%d %s" % (n,names[c]) for (n,c) in comb))
        return 1
    cur = denominations[i]
    return sum(count_combs(left-x*cur, i+1, comb[:], (x,cur)) for x in range(0, int(left/cur)+1))

count_combs(cents, 0, [], None)

Answer 4

Scala函数：

def countChange(money: Int, coins: List[Int]): Int = {

def loop(money: Int, lcoins: List[Int], count: Int): Int = {
  // if there are no more coins or if we run out of money ... return 0 
  if ( lcoins.isEmpty || money < 0) 0
  else{
    if (money == 0 ) count + 1   
/* if the recursive subtraction leads to 0 money left - a prefect division hence return count +1 */
    else
/* keep iterating ... sum over money and the rest of the coins and money - the first item and the full set of coins left*/
      loop(money, lcoins.tail,count) + loop(money - lcoins.head,lcoins, count)
  }
}

val x = loop(money, coins, 0)
Console println x
x
}

Answer 5

这里有一些绝对简单的C ++代码来解决问题，它确实要求显示所有组合。

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

int main(int argc, char *argv[])
{
    if (argc != 2)
    {
        printf("usage: change amount-in-cents\n");
        return 1;
    }

    int total = atoi(argv[1]);

    printf("quarter\tdime\tnickle\tpenny\tto make %d\n", total);

    int combos = 0;

    for (int q = 0; q <= total / 25; q++)
    {
        int total_less_q = total - q * 25;
        for (int d = 0; d <= total_less_q / 10; d++)
        {
            int total_less_q_d = total_less_q - d * 10;
            for (int n = 0; n <= total_less_q_d / 5; n++)
            {
                int p = total_less_q_d - n * 5;
                printf("%d\t%d\t%d\t%d\n", q, d, n, p);
                combos++;
            }
        }
    }

    printf("%d combinations\n", combos);

    return 0;
}

但我对仅仅计算组合数的子问题非常感兴趣。我怀疑它有一个封闭形式的等式。

Answer 6

代码使用Java来解决这个问题，它也可以运行......由于循环太多，这个方法可能不是一个好主意，但它确实是一种直接的方式。

public class RepresentCents {

    public static int sum(int n) {

        int count = 0;
        for (int i = 0; i <= n / 25; i++) {
            for (int j = 0; j <= n / 10; j++) {
                for (int k = 0; k <= n / 5; k++) {
                    for (int l = 0; l <= n; l++) {
                        int v = i * 25 + j * 10 + k * 5 + l;
                        if (v == n) {
                            count++;
                        } else if (v > n) {
                            break;
                        }
                    }
                }
            }
        }
        return count;
    }

    public static void main(String[] args) {
        System.out.println(sum(100));
    }
}

Answer 7

这是一个非常古老的问题，但我想出了一个java中的递归解决方案，似乎比其他所有解决方案都要小，所以这里有 -

 public static void printAll(int ind, int[] denom,int N,int[] vals){
    if(N==0){
        System.out.println(Arrays.toString(vals));
        return;
    }
    if(ind == (denom.length))return;             
    int currdenom = denom[ind];
    for(int i=0;i<=(N/currdenom);i++){
        vals[ind] = i;
        printAll(ind+1,denom,N-i*currdenom,vals);
    }
 }

改进：

  public static void printAllCents(int ind, int[] denom,int N,int[] vals){
        if(N==0){
            if(ind < denom.length) {
                for(int i=ind;i<denom.length;i++)
                    vals[i] = 0;
            }
            System.out.println(Arrays.toString(vals));
            return;
        }
        if(ind == (denom.length)) {
            vals[ind-1] = 0;
            return;             
        }

        int currdenom = denom[ind];
        for(int i=0;i<=(N/currdenom);i++){ 
                vals[ind] = i;
                printAllCents(ind+1,denom,N-i*currdenom,vals);
        }
     }

Answer 8

子问题是典型的动态规划问题。

/* Q: Given some dollar value in cents (e.g. 200 = 2 dollars, 1000 = 10 dollars),
      find the number of combinations of coins that make up the dollar value.
      There are only penny, nickel, dime, and quarter.
      (quarter = 25 cents, dime = 10 cents, nickel = 5 cents, penny = 1 cent) */
/* A:
Reference: http://andrew.neitsch.ca/publications/m496pres1.nb.pdf
f(n, k): number of ways of making change for n cents, using only the first
         k+1 types of coins.

          +- 0,                        n < 0 || k < 0
f(n, k) = |- 1,                        n == 0
          +- f(n, k-1) + f(n-C[k], k), else
 */

#include <iostream>
#include <vector>
using namespace std;

int C[] = {1, 5, 10, 25};

// Recursive: very slow, O(2^n)
int f(int n, int k)
{
    if (n < 0 || k < 0)
        return 0;

    if (n == 0)
        return 1;

    return f(n, k-1) + f(n-C[k], k); 
}

// Non-recursive: fast, but still O(nk)
int f_NonRec(int n, int k)
{
    vector<vector<int> > table(n+1, vector<int>(k+1, 1));

    for (int i = 0; i <= n; ++i)
    {
        for (int j = 0; j <= k; ++j)
        {
            if (i < 0 || j < 0) // Impossible, for illustration purpose
            {
                table[i][j] = 0;
            }
            else if (i == 0 || j == 0) // Very Important
            {
                table[i][j] = 1;
            }
            else
            {
                // The recursion. Be careful with the vector boundary
                table[i][j] = table[i][j-1] + 
                    (i < C[j] ? 0 : table[i-C[j]][j]);
            }
        }
    }

    return table[n][k];
}

int main()
{
    cout << f(100, 3) << ", " << f_NonRec(100, 3) << endl;
    cout << f(200, 3) << ", " << f_NonRec(200, 3) << endl;
    cout << f(1000, 3) << ", " << f_NonRec(1000, 3) << endl;

    return 0;
}

Answer 9

设C（i，J）使用集合J中的值制作i美分的组合。

您可以将C定义为：

（first（J）以确定的方式 集合的元素）

事实证明这是一个非常递归的函数......如果你使用memoization，它会相当有效;）

Answer 10

半黑客解决这个独特的组合问题 - 强制降序：

$denoms = [1,5,10,25]
def all_combs(sum,last) 
  return 1 if sum == 0
  return $denoms.select{|d| d &le sum && d &le last}.inject(0) {|total,denom|
           total+all_combs(sum-denom,denom)}
end

这会慢，因为它不会被记忆，但你明白了。

Answer 11

# short and sweet with O(n) table memory    

#include <iostream>
#include <vector>

int count( std::vector<int> s, int n )
{
  std::vector<int> table(n+1,0);

  table[0] = 1;
  for ( auto& k : s )
    for(int j=k; j<=n; ++j)
      table[j] += table[j-k];

  return table[n];
}

int main()
{
  std::cout <<  count({25, 10, 5, 1}, 100) << std::endl;
  return 0;
}

Answer 12

这是我在Python中的答案。它不使用递归：

def crossprod (list1, list2):
    output = 0
    for i in range(0,len(list1)):
        output += list1[i]*list2[i]

    return output

def breakit(target, coins):
    coinslimit = [(target / coins[i]) for i in range(0,len(coins))]
    count = 0
    temp = []
    for i in range(0,len(coins)):
        temp.append([j for j in range(0,coinslimit[i]+1)])


    r=[[]]
    for x in temp:
        t = []
        for y in x:
            for i in r:
                t.append(i+[y])
        r = t

    for targets in r:
        if crossprod(targets, coins) == target:
            print targets
            count +=1
    return count




if __name__ == "__main__":
    coins = [25,10,5,1]
    target = 78
    print breakit(target, coins)

示例输出

    ...
    1 ( 10 cents)  2 ( 5 cents)  58 ( 1 cents)  
    4 ( 5 cents)  58 ( 1 cents)  
    1 ( 10 cents)  1 ( 5 cents)  63 ( 1 cents)  
    3 ( 5 cents)  63 ( 1 cents)  
    1 ( 10 cents)  68 ( 1 cents)  
    2 ( 5 cents)  68 ( 1 cents)  
    1 ( 5 cents)  73 ( 1 cents)  
    78 ( 1 cents)  
    Number of solutions =  121

Answer 13

var countChange = function (money,coins) {
  function countChangeSub(money,coins,n) {
    if(money==0) return 1;
    if(money<0 || coins.length ==n) return 0;
    return countChangeSub(money-coins[n],coins,n) + countChangeSub(money,coins,n+1);
  }
  return countChangeSub(money,coins,0);
}

Answer 14

我知道这是一个非常古老的问题。我正在寻找正确的答案，无法找到任何简单而令人满意的东西。花了我一些时间，但能够记下一些东西。

function denomination(coins, original_amount){
    var original_amount = original_amount;
    var original_best = [ ];

    for(var i=0;i<coins.length; i++){
      var amount = original_amount;
      var best = [ ];
      var tempBest = [ ]
      while(coins[i]<=amount){
        amount = amount - coins[i];
        best.push(coins[i]);
      }
      if(amount>0 && coins.length>1){
        tempBest = denomination(coins.slice(0,i).concat(coins.slice(i+1,coins.length)), amount);
        //best = best.concat(denomination(coins.splice(i,1), amount));
      }
      if(tempBest.length!=0 || (best.length!=0 && amount==0)){
        best = best.concat(tempBest);
        if(original_best.length==0 ){
          original_best = best
        }else if(original_best.length > best.length ){
          original_best = best;
        }  
      }
    }
    return original_best;  
  }
  denomination( [1,10,3,9] , 19 );

这是一个javascript解决方案并使用递归。

Answer 15

如果货币系统允许，只需一个简单的greedy algorithm，从最高价值的货币开始，尽可能多地获取每枚硬币。

否则，需要动态编程才能快速找到最佳解决方案，因为此问题基本上是knapsack problem。

例如，如果货币系统有硬币：{13, 8, 1}，则贪婪的解决方案会将24更改为{13, 8, 1, 1, 1}，但真正的最佳解决方案是{8, 8, 8}

编辑：我认为我们正在以最佳方式进行变革，而不是列出所有改变美元的方法。我最近的采访询问如何进行改变，所以我在完成阅读问题之前就跳了起来。

Answer 16

两者：迭代所有面额从高到低，取一个面额，减去所需的总数，然后递归余数（将可用的面额限制为等于或低于当前迭代值。）

Answer 17

在Scala编程语言中，我会这样做：

 def countChange(money: Int, coins: List[Int]): Int = {

       money match {
           case 0 => 1
           case x if x < 0 => 0
           case x if x >= 1 && coins.isEmpty => 0
           case _ => countChange(money, coins.tail) + countChange(money - coins.head, coins)

       }

  }

Answer 18

This blog entry of mine为XKCD comic中的数字解决了这个背包问题。对items dict和exactcost值进行简单更改也会为您的问题提供所有解决方案。

如果问题是找到使用成本最低的变化，那么使用尽可能多的最高价值硬币的天真贪婪算法可能会因某些硬币和目标金额组合而失败。例如，如果有值为1,3和4的硬币;当目标数量为6时，贪婪算法可能会建议三个值为4,1和1的硬币，因为很容易看出你可以使用两个价值为3的硬币。

• 水稻

Answer 19

这是一个简单的递归算法，它接受一个账单，然后递归地收取较小的账单，直到它达到总和，然后再拿另一个相同面额的账单，并再次递归。请参阅下面的示例输出以进行说明。

var bills = new int[] { 100, 50, 20, 10, 5, 1 };

void PrintAllWaysToMakeChange(int sumSoFar, int minBill, string changeSoFar)
{
    for (int i = minBill; i < bills.Length; i++)
    {
        var change = changeSoFar;
        var sum = sumSoFar;

        while (sum > 0)
        {
            if (!string.IsNullOrEmpty(change)) change += " + ";
            change += bills[i];

            sum -= bills[i]; 
            if (sum > 0)
            {
                PrintAllWaysToMakeChange(sum, i + 1, change);
            }
        }

        if (sum == 0)
        {
            Console.WriteLine(change);
        }
    }
}

PrintAllWaysToMakeChange(15, 0, "");

打印以下内容：

10 + 5
10 + 1 + 1 + 1 + 1 + 1
5 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
5 + 5 + 1 + 1 + 1 + 1 + 1
5 + 5 + 5
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

Answer 20

直截了当的java解决方案：

public static void main(String[] args) 
{    
    int[] denoms = {4,2,3,1};
    int[] vals = new int[denoms.length];
    int target = 6;
    printCombinations(0, denoms, target, vals);
}


public static void printCombinations(int index, int[] denom,int target, int[] vals)
{
  if(target==0)
  {
    System.out.println(Arrays.toString(vals));
    return;
  }
  if(index == denom.length) return;   
  int currDenom = denom[index];
  for(int i = 0; i*currDenom <= target;i++)
  {
    vals[index] = i;
    printCombinations(index+1, denom, target - i*currDenom, vals);
    vals[index] = 0;
  }
}

Answer 21

这是Zihan答案的改进。当面额仅为1美分时，会产生大量不必要的循环。

直观且非递归。

    public static int Ways2PayNCents(int n)
    {
        int numberOfWays=0;
        int cent, nickel, dime, quarter;
        for (quarter = 0; quarter <= n/25; quarter++)
        {
            for (dime = 0; dime <= n/10; dime++)
            {
                for (nickel = 0; nickel <= n/5; nickel++)
                {
                    cent = n - (quarter * 25 + dime * 10 + nickel * 5);
                    if (cent >= 0)
                    {
                        numberOfWays += 1;
                        Console.WriteLine("{0},{1},{2},{3}", quarter, dime, nickel, cent);
                    }                   
                }
            }
        }
        return numberOfWays;            
    }

Answer 22

/*
* make a list of all distinct sets of coins of from the set of coins to
* sum up to the given target amount.
* Here the input set of coins is assumed yo be {1, 2, 4}, this set MUST
* have the coins sorted in ascending order.
* Outline of the algorithm:
* 
* Keep track of what the current coin is, say ccn; current number of coins
* in the partial solution, say k; current sum, say sum, obtained by adding
* ccn; sum sofar, say accsum:
*  1) Use ccn as long as it can be added without exceeding the target
*     a) if current sum equals target, add cc to solution coin set, increase
*     coin coin in the solution by 1, and print it and return
*     b) if current sum exceeds target, ccn can't be in the solution, so
*        return
*     c) if neither of the above, add current coin to partial solution,
*        increase k by 1 (number of coins in partial solution), and recuse
*  2) When current denomination can no longer be used, start using the
*     next higher denomination coins, just like in (1)
*  3) When all denominations have been used, we are done
*/

#include <iostream>
#include <cstdlib>

using namespace std;

// int num_calls = 0;
// int num_ways = 0;

void print(const int coins[], int n);

void combine_coins(
                   const int denoms[], // coins sorted in ascending order
                   int n,              // number of denominations
                   int target,         // target sum
                   int accsum,         // accumulated sum
                   int coins[],        // solution set, MUST equal
                                       // target / lowest denom coin
                   int k               // number of coins in coins[]
                  )
{

    int  ccn;   // current coin
    int  sum;   // current sum

    // ++num_calls;

    for (int i = 0; i < n; ++i) {
        /*
         * skip coins of lesser denomination: This is to be efficient
         * and also avoid generating duplicate sequences. What we need
         * is combinations and without this check we will generate
         * permutations.
         */
        if (k > 0 && denoms[i] < coins[k - 1])
            continue;   // skip coins of lesser denomination

        ccn = denoms[i];

        if ((sum = accsum + ccn) > target)
            return;     // no point trying higher denominations now


        if (sum == target) {
            // found yet another solution
            coins[k] = ccn;
            print(coins, k + 1);
            // ++num_ways;
            return;
        }

        coins[k] = ccn;
        combine_coins(denoms, n, target, sum, coins, k + 1);
    }
}

void print(const int coins[], int n)
{
    int s = 0;
    for (int i = 0; i < n; ++i) {
        cout << coins[i] << " ";
        s += coins[i];
    }
    cout << "\t = \t" << s << "\n";

}

int main(int argc, const char *argv[])
{

    int denoms[] = {1, 2, 4};
    int dsize = sizeof(denoms) / sizeof(denoms[0]);
    int target;

    if (argv[1])
        target = atoi(argv[1]);
    else
        target = 8;

    int *coins = new int[target];


    combine_coins(denoms, dsize, target, 0, coins, 0);

    // cout << "num calls = " << num_calls << ", num ways = " << num_ways << "\n";

    return 0;
}

Answer 23

这是一个C＃函数：

    public static void change(int money, List<int> coins, List<int> combination)
    {
        if(money < 0 || coins.Count == 0) return;
        if (money == 0)
        {
            Console.WriteLine((String.Join("; ", combination)));
            return;
        }

        List<int> copy = new List<int>(coins);
        copy.RemoveAt(0);
        change(money, copy, combination);

        combination = new List<int>(combination) { coins[0] };
        change(money - coins[0], coins, new List<int>(combination));
    }

像这样使用：

change(100, new List<int>() {5, 10, 25}, new List<int>());

打印：

25; 25; 25; 25
10; 10; 10; 10; 10; 25; 25
10; 10; 10; 10; 10; 10; 10; 10; 10; 10
5; 10; 10; 25; 25; 25
5; 10; 10; 10; 10; 10; 10; 10; 25
5; 5; 10; 10; 10; 10; 25; 25
5; 5; 10; 10; 10; 10; 10; 10; 10; 10; 10
5; 5; 5; 10; 25; 25; 25
5; 5; 5; 10; 10; 10; 10; 10; 10; 25
5; 5; 5; 5; 10; 10; 10; 25; 25
5; 5; 5; 5; 10; 10; 10; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 25; 25; 25
5; 5; 5; 5; 5; 10; 10; 10; 10; 10; 25
5; 5; 5; 5; 5; 5; 10; 10; 25; 25
5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 10; 25; 25
5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 25; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 25
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 10
5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5; 5

Answer 24

我在O'reily的“Python For Data Analysis”一书中找到了这段简洁的代码。它使用延迟实现和int比较，我认为它可以使用小数修改为其他面额。让我知道它对你有用！

def make_change(amount, coins=[1, 5, 10, 25], hand=None):
 hand = [] if hand is None else hand
 if amount == 0:
 yield hand
 for coin in coins:
 # ensures we don't give too much change, and combinations are unique
 if coin > amount or (len(hand) > 0 and hand[-1] < coin):
 continue
 for result in make_change(amount - coin, coins=coins,
 hand=hand + [coin]):
 yield result

Answer 25

public class Coins {

static int ac = 421;
static int bc = 311;
static int cc = 11;

static int target = 4000;

public static void main(String[] args) {


    method2();
}

  public static void method2(){
    //running time n^2

    int da = target/ac;
    int db = target/bc;     

    for(int i=0;i<=da;i++){         
        for(int j=0;j<=db;j++){             
            int rem = target-(i*ac+j*bc);               
            if(rem < 0){                    
                break;                  
            }else{                  
                if(rem%cc==0){                  
                    System.out.format("\n%d, %d, %d ---- %d + %d + %d = %d \n", i, j, rem/cc, i*ac, j*bc, (rem/cc)*cc, target);                     
                }                   
            }                   
        }           
    }       
}
 }

Answer 26

Duh，我现在觉得很蠢。下面是一个过于复杂的解决方案，我会保留它，因为它毕竟是 解决方案。一个简单的解决方案就是：

// Generate a pretty string
val coinNames = List(("quarter", "quarters"), 
                     ("dime", "dimes"), 
                     ("nickel", "nickels"), 
                     ("penny", "pennies"))
def coinsString = 
  Function.tupled((quarters: Int, dimes: Int, nickels:Int, pennies: Int) => (
    List(quarters, dimes, nickels, pennies) 
    zip coinNames // join with names
    map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
    map (t => t._1 + " " + t._2) // qty name
    mkString " "
  ))

def allCombinations(amount: Int) = 
 (for{quarters <- 0 to (amount / 25)
      dimes <- 0 to ((amount - 25*quarters) / 10)
      nickels <- 0 to ((amount - 25*quarters - 10*dimes) / 5)
  } yield (quarters, dimes, nickels, amount - 25*quarters - 10*dimes - 5*nickels)
 ) map coinsString mkString "\n"

这是另一种解决方案。该解决方案基于观察到每个硬币是其他硬币的倍数，因此它们可以用它们来表示。

// Just to make things a bit more readable, as these routines will access
// arrays a lot
val coinValues = List(25, 10, 5, 1)
val coinNames = List(("quarter", "quarters"), 
                     ("dime", "dimes"), 
                     ("nickel", "nickels"), 
                     ("penny", "pennies"))
val List(quarter, dime, nickel, penny) = coinValues.indices.toList


// Find the combination that uses the least amount of coins
def leastCoins(amount: Int): Array[Int] =
  ((List(amount) /: coinValues) {(list, coinValue) =>
    val currentAmount = list.head
    val numberOfCoins = currentAmount / coinValue
    val remainingAmount = currentAmount % coinValue
    remainingAmount :: numberOfCoins :: list.tail
  }).tail.reverse.toArray

// Helper function. Adjust a certain amount of coins by
// adding or subtracting coins of each type; this could
// be made to receive a list of adjustments, but for so
// few types of coins, it's not worth it.
def adjust(base: Array[Int], 
           quarters: Int, 
           dimes: Int, 
           nickels: Int, 
           pennies: Int): Array[Int] =
  Array(base(quarter) + quarters, 
        base(dime) + dimes, 
        base(nickel) + nickels, 
        base(penny) + pennies)

// We decrease the amount of quarters by one this way
def decreaseQuarter(base: Array[Int]): Array[Int] =
  adjust(base, -1, +2, +1, 0)

// Dimes are decreased this way
def decreaseDime(base: Array[Int]): Array[Int] =
  adjust(base, 0, -1, +2, 0)

// And here is how we decrease Nickels
def decreaseNickel(base: Array[Int]): Array[Int] =
  adjust(base, 0, 0, -1, +5)

// This will help us find the proper decrease function
val decrease = Map(quarter -> decreaseQuarter _,
                   dime -> decreaseDime _,
                   nickel -> decreaseNickel _)

// Given a base amount of coins of each type, and the type of coin,
// we'll produce a list of coin amounts for each quantity of that particular
// coin type, up to the "base" amount
def coinSpan(base: Array[Int], whichCoin: Int) = 
  (List(base) /: (0 until base(whichCoin)).toList) { (list, _) =>
    decrease(whichCoin)(list.head) :: list
  }

// Generate a pretty string
def coinsString(base: Array[Int]) = (
  base 
  zip coinNames // join with names
  map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
  map (t => t._1 + " " + t._2)
  mkString " "
)

// So, get a base amount, compute a list for all quarters variations of that base,
// then, for each combination, compute all variations of dimes, and then repeat
// for all variations of nickels.
def allCombinations(amount: Int) = {
  val base = leastCoins(amount)
  val allQuarters = coinSpan(base, quarter)
  val allDimes = allQuarters flatMap (base => coinSpan(base, dime))
  val allNickels = allDimes flatMap (base => coinSpan(base, nickel))
  allNickels map coinsString mkString "\n"
}

因此，对于37个硬币，例如：

scala> println(allCombinations(37))
0 quarter 0 dimes 0 nickels 37 pennies
0 quarter 0 dimes 1 nickel 32 pennies
0 quarter 0 dimes 2 nickels 27 pennies
0 quarter 0 dimes 3 nickels 22 pennies
0 quarter 0 dimes 4 nickels 17 pennies
0 quarter 0 dimes 5 nickels 12 pennies
0 quarter 0 dimes 6 nickels 7 pennies
0 quarter 0 dimes 7 nickels 2 pennies
0 quarter 1 dime 0 nickels 27 pennies
0 quarter 1 dime 1 nickel 22 pennies
0 quarter 1 dime 2 nickels 17 pennies
0 quarter 1 dime 3 nickels 12 pennies
0 quarter 1 dime 4 nickels 7 pennies
0 quarter 1 dime 5 nickels 2 pennies
0 quarter 2 dimes 0 nickels 17 pennies
0 quarter 2 dimes 1 nickel 12 pennies
0 quarter 2 dimes 2 nickels 7 pennies
0 quarter 2 dimes 3 nickels 2 pennies
0 quarter 3 dimes 0 nickels 7 pennies
0 quarter 3 dimes 1 nickel 2 pennies
1 quarter 0 dimes 0 nickels 12 pennies
1 quarter 0 dimes 1 nickel 7 pennies
1 quarter 0 dimes 2 nickels 2 pennies
1 quarter 1 dime 0 nickels 2 pennies

Answer 27

下面是一个Python程序，用于查找所有货币组合。这是一个具有order（n）时间的动态编程解决方案。 钱是1,5,10,25

我们从第1行货币遍历到第25行货币（4行）。如果我们仅考虑货币1，则行货币1包含计数 计算组合数。行货币5通过取行货币r的计数来产生每一列 相同的最终货币加上其自己行中的前5个计数（当前头寸减去5）。行货币10使用行货币5， 其中包含两个1,5的计数，并加上前10个计数（当前位置减去10）。行钱25使用行 货币10，其中包含行货币1,5,10的计数加上前25个计数。

例如，数字[1] [12] =数字[0] [12] +数字[1] [7]（7 = 12-5），结果为3 = 1 + 2；数字[3] [12] = 数字[2] [12] +数字[3] [9]（-13 = 12-25），结果是4 = 0 + 4，因为-13小于0。

def cntMoney(num):
    mSz = len(money)
    numbers = [[0]*(1+num) for _ in range(mSz)]
    for mI in range(mSz): numbers[mI][0] = 1
    for mI,m in enumerate(money):
        for i in range(1,num+1):
            numbers[mI][i] = numbers[mI][i-m] if i >= m else 0
            if mI != 0: numbers[mI][i] += numbers[mI-1][i]
        print('m,numbers',m,numbers[mI])
    return numbers[mSz-1][num]

money = [1,5,10,25]
    num = 12
    print('money,combinations',num,cntMoney(num))

output:    
('m,numbers', 1, [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])
('m,numbers', 5, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3])
('m,numbers', 10, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('m,numbers', 25, [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 4])
('money,combinations', 12, 4)

Answer 28

Mathematica 中的以下单行代码产生了想要的结果

parts[n_]:=Flatten[Table[{(n-i)/25,(i-j)/10,(j-k)/5,k},{i,n,0,-25},{j,i,0,-10},{k,j,0,-5}],2]

该表通过迭代有多少较大种类的硬币适合 n 的所有方式来简单地建立所有组合，同时对剩余的较小种类的硬币重复相同的步骤。该实现在任何其他编程语言中都应该是类似的。

输出是带有符号的元素列表：

{number of quarters, number of dimes, number of nickels, number of cents}

结构背后的直觉可以从一个例子中读出：

parts[51]

<块引用>

Answer 29

这里有很多变化，但找不到任何地方组合数量的PHP解决方案，所以我会添加一个。

/**
 * @param int $money The total value
 * @param array $coins The coin denominations
 * @param int $sum The countable sum
 * @return int
 */
function getTotalCombinations($money, $coins, &$sum = 0){
  if ($money == 0){
    return $sum++;
  } else if (empty($coins) || $money < 0){
    return $sum;
  } else {
      $firstCoin = array_pop(array_reverse($coins));
      getTotalCombinations($money - $firstCoin, $coins, $sum) + getTotalCombinations($money, array_diff($coins, [$firstCoin]), $sum);
  }
  return $sum;
}


$totalCombinations = getTotalCombinations($money, $coins);

Answer 30

我在 C# 中实现了这个谜题，我认为它与其他答案有点不同。此外，我添加了注释以使我的算法更易于理解。这是一个非常好的拼图。

从较大的硬币开始，查看所有 Q,D,N 并用便士填满其余部分。

所以我可以从 0 四分之一、0 角钱和 0 镍开始，其余的都是便士

对于每个硬币，我有 ($Amount / $Change) + 1，其中 $Amount 是输入参数，$Change 是 [Q]uarter 或 [D]ime 或 [N]ickel。所以，假设输入参数是$1，

• 对于 Quarter (0, 1, 2, 3, 4) - 我的代码中有 ($1 / 0.25) + 1 = 5 个选项 -- qLoop 变量
• 对于相同的 $1，我将有 ($1 / 0.10) + 1 = 11 种 Dime (0, 1, 2 ,3, 4, 5, 6, 7, 8, 9, 10) -- {{ 1}} 我代码中的变量
• 对于 Nickel，我将有 ($1 / 0.05) + 1 = 21 选项（0 到 20） -- dLoop 变量在我的代码中

请注意，在每个循环中，我必须使用来自外循环的提醒。

举个例子，如果我选择 1 个季度（其中 q 是 1），我还有 $1 - 0.25 用于内部循环（Dime 循环）等等

nLoop

输出

Answer 31

PHP代码将金额细分为面额：

//Define the denominations    
private $denominations = array(1000, 500, 200, 100, 50, 40, 20, 10, 5, 1);
/**
 * S# countDenomination() function
 * 
 * @author Edwin Mugendi <edwinmugendi@gmail.com>
 * 
 * Count denomination
 * 
 * @param float $original_amount Original amount
 * 
 * @return array with denomination and count
 */
public function countDenomination($original_amount) {
    $amount = $original_amount;
    $denomination_count_array = array();
    foreach ($this->denominations as $single_denomination) {

        $count = floor($amount / $single_denomination);

        $denomination_count_array[$single_denomination] = $count;

        $amount = fmod($amount, $single_denomination);
    }//E# foreach statement

    var_dump($denomination_count_array);
    return $denomination_count_array;
    //return $denomination_count_array;
}

// E＃countDenomination（）函数

Answer 32

下面是python解决方案：

    x = []
    dic = {}
    def f(n,r):
        [a,b,c,d] = r
        if not dic.has_key((n,a,b,c,d)): dic[(n,a,b,c,d)] = 1
        if n>=25:
            if not dic.has_key((n-25,a+1,b,c,d)):f(n-25,[a+1,b,c,d])
            if not dic.has_key((n-10,a,b+1,c,d)):f(n-10,[a,b+1,c,d])
            if not dic.has_key((n-5,a,b,c+1,d)):f(n-5,[a,b,c+1,d])
            if not dic.has_key((n-1,a,b,c,d+1)):f(n-1,[a,b,c,d+1])
        elif n>=10:
            if not dic.has_key((n-10,a,b+1,c,d)):f(n-10,[a,b+1,c,d])
            if not dic.has_key((n-5,a,b,c+1,d)):f(n-5,[a,b,c+1,d])
            if not dic.has_key((n-1,a,b,c,d+1)):f(n-1,[a,b,c,d+1])
        elif n>=5:
            if not dic.has_key((n-5,a,b,c+1,d)):f(n-5,[a,b,c+1,d])
            if not dic.has_key((n-1,a,b,c,d+1)):f(n-1,[a,b,c,d+1])
        elif n>=1:
            if not dic.has_key((n-1,a,b,c,d+1)):f(n-1,[a,b,c,d+1])
        else:
            if r not in x:
                x.extend([r])

    f(100, [0,0,0,0])
    print x

Answer 33

这是一个基于python的解决方案，它使用递归和memoization导致复杂度为O（mxn）

    def get_combinations_dynamic(self, amount, coins, memo):
    end_index = len(coins) - 1
    memo_key = str(amount)+'->'+str(coins)
    if memo_key in memo:
        return memo[memo_key]
    remaining_amount = amount
    if amount < 0:
        return []
    if amount == 0:
        return [[]]
    combinations = []
    if len(coins) <= 1:
        if amount % coins[0] == 0:
            combination = []
            for i in range(amount // coins[0]):
                combination.append(coins[0])
            list.sort(combination)
            if combination not in combinations:
                combinations.append(combination)
    else:
        k = 0
        while remaining_amount >= 0:
            sub_combinations = self.get_combinations_dynamic(remaining_amount, coins[:end_index], memo)
            for combination in sub_combinations:
                temp = combination[:]
                for i in range(k):
                    temp.append(coins[end_index])
                list.sort(temp)
                if temp not in combinations:
                    combinations.append(temp)
            k += 1
            remaining_amount -= coins[end_index]
    memo[memo_key] = combinations
    return combinations

Answer 34

以下java解决方案也将打印不同的组合。容易明白。想法是

总和5

解决方案是

    5 - 5(i) times 1 = 0
        if(sum = 0)
           print i times 1
    5 - 4(i) times 1 = 1
    5 - 3 times 1 = 2
        2 -  1(j) times 2 = 0
           if(sum = 0)
              print i times 1 and j times 2
    and so on......

如果每个循环中的剩余总和小于面额，即 如果剩余和1小于2，则只需打破循环

以下完整代码

如有任何错误，请纠正我

public class CoinCombinbationSimple {
public static void main(String[] args) {
    int sum = 100000;
    printCombination(sum);
}

static void printCombination(int sum) {
    for (int i = sum; i >= 0; i--) {
        int sumCopy1 = sum - i * 1;
        if (sumCopy1 == 0) {
            System.out.println(i + " 1 coins");
        }
        for (int j = sumCopy1 / 2; j >= 0; j--) {
            int sumCopy2 = sumCopy1;
            if (sumCopy2 < 2) {
                break;
            }
            sumCopy2 = sumCopy1 - 2 * j;
            if (sumCopy2 == 0) {
                System.out.println(i + " 1 coins " + j + " 2 coins ");
            }
            for (int k = sumCopy2 / 5; k >= 0; k--) {
                int sumCopy3 = sumCopy2;
                if (sumCopy2 < 5) {
                    break;
                }
                sumCopy3 = sumCopy2 - 5 * k;
                if (sumCopy3 == 0) {
                    System.out.println(i + " 1 coins " + j + " 2 coins "
                            + k + " 5 coins");
                }
            }
        }
    }
}

}

Answer 35

Java解决方案

import java.util.Arrays;
import java.util.Scanner;


public class nCents {



public static void main(String[] args) {

    Scanner input=new Scanner(System.in);
    int cents=input.nextInt();
    int num_ways [][] =new int [5][cents+1];

    //putting in zeroes to offset
    int getCents[]={0 , 0 , 5 , 10 , 25};
    Arrays.fill(num_ways[0], 0);
    Arrays.fill(num_ways[1], 1);

    int current_cent=0;
    for(int i=2;i<num_ways.length;i++){

        current_cent=getCents[i];

        for(int j=1;j<num_ways[0].length;j++){
            if(j-current_cent>=0){
                if(j-current_cent==0){
                    num_ways[i][j]=num_ways[i-1][j]+1;
                }else{
                    num_ways[i][j]=num_ways[i][j-current_cent]+num_ways[i-1][j];
                }
            }else{
                num_ways[i][j]=num_ways[i-1][j];
            }


        }


    }



    System.out.println(num_ways[num_ways.length-1][num_ways[0].length-1]);

}

}

Answer 36

我使用一个非常简单的循环来解决这个问题，我正在使用Isogenic Game Engine在HTML5中编写BlackJack游戏。您可以看到BlackJack游戏的视频，该视频显示了用于在牌上方BlackJack牌桌上的投注值进行投注的筹码：http://bit.ly/yUF6iw

在此示例中，betValue等于您希望分为“硬币”或“筹码”等的总值。

您可以将chipValues数组项设置为您的硬币或筹码的价值。确保从最低值到最高值（便士，镍，角钱，四分之一）订购物品。

这是JavaScript：

// Set the total that we want to divide into chips
var betValue = 191;

// Set the chip values
var chipValues = [
    1,
    5,
    10,
    25
];

// Work out how many of each chip is required to make up the bet value
var tempBet = betValue;
var tempChips = [];
for (var i = chipValues.length - 1; i >= 0; i--) {
    var chipValue = chipValues[i];
    var divided = Math.floor(tempBet / chipValue);

    if (divided >= 1) {
        tempChips[i] = divided;
        tempBet -= divided * chipValues[i];
    }

    if (tempBet == 0) { break; }
}

// Display the chips and how many of each make up the betValue
for (var i in tempChips) {
    console.log(tempChips[i] + ' of ' + chipValues[i]);
}

你显然不需要做最后一个循环，它只是在console.log中的最终数组值。

Answer 37

public static int calcCoins(int cents){
        return coins(cents,new int[cents+1]);
    }
    public static int coins(int cents,int[] memo){
        if(memo[cents] != 0){
            return -1;
        }
        int sum = 0;
        int arr[] = {25,10,5,1};

        for (int i = 0; i < arr.length; i++) {
            if(cents/arr[i] != 0){
                int temp = coins(cents-arr[i],memo);
                if(temp == 0){
                    sum+=1;
                } else if(temp == -1){
                    sum +=0;
                }
                else{
                    sum += temp;
                }
            } 
        }
        memo[cents] = sum;
        return sum;
    }