硬币组合问题

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题设：

假设我们有8种不同面值的硬币｛1，2，5，10，20，50，100｝，用这些硬币组合够成一个给定的数值n。例如n=200，那么一种可能的组合方式为 200 = 3 * 1 + 1＊2 + 1＊5 + 2＊20 + 1 * 50 + 1 * 100。问总过有多少种可能的组合方式？

思路分析：

sum = x1 * V1 + x2 * V2 + ... + 0 * Vm

sum = x1 * V1 + x2 * V2 + ... + 1 * Vm

sum = x1 * V1 + x2 * V2 + ... + 2 * Vm

...

sum = x1 * V1 + x2 * V2 + ... + K * Vm

(V1,V2,...,Vm)代表硬币值，其中K是该xm能取的最大数值K = sum / Vm。

可以递推出如下公式：

dp[i][sum] = dp[i-1][sum - 0*Vm] + dp[i-1][sum - 1*Vm+

dp[i-1][sum - 2*Vm] + ... + dp[i-1][sum - K*Vm]

初始值设定：

如果sum=0，那么无论有前多少种来组合0，只有一种可能，就是各个系数都等于0；

dp[i][0] = 1 // i = 0, 1, 2, ... , m

　　如果我们用二位数组表示dp[i][sum], 我们发现第i行的值全部依赖与i-1行的值，所以我们可以逐行求解该数组。如果前0种硬币要组成sum，我们规定为dp[0][sum] = 0。

代码如下：

/*  2  * Filename :coins.cpp 3  * Description: solve coin combinations using dynamic programing 4  * Complier: g++ 5  * Author: python27 6  */ #include <iostream> #include <string> #include <cmath> #include <vector> using namespace std;  /****************************************************************  * coin Combinations: using dynamic programming  *  * Basic idea:  * dp[i][j] = sum(dp[i-1][j-k*coins[i-1]]) for k = 1,2,..., j/coins[i-1]  * dp[0][j] = 1 for j = 0, 1, 2, ..., sum  *   * Input:  * coins[] - array store all values of the coins  * coinKinds - how many kinds of coins there are  * sum - the number you want to construct using coins  *  * Output:  * the number of combinations using coins construct sum  *  * Usage:  * c[3] = {1, 2, 5};  * int result = coinCombinations(c, 3, 10);  *  ****************************************************************/ int coinCombinations(int coins[], int coinKinds, int sum) {    // 2-D array using vector: is equal to: dp[coinKinds+1][sum+1] = {0};     vector<vector<int> > dp(coinKinds + 1);     for (int i = 0; i <= coinKinds; ++i)    {        dp[i].resize(sum + 1);   //reset dp[i] to the size of sum+1    }    /* for (int i = 0; i <= coinKinds; ++i)     {         for (int j = 0; j <= sum; ++j)         {             dp[i][j] = 0;         }     }*/      //init: dp[i][0] = 1; i = 0, 1, 2 ..., coinKinds      //Notice: dp[0][0] must be 1, althongh it make no sense that      //using 0 kinds of coins construct 0 has one way. but it is the foundation      //of iteration. without it everything based on it goes wrong    for (int i = 0; i <= coinKinds; ++i)    {        dp[i][0] = 1;    }for(int j = 1;j <= sum; j++){dp[0][j] = 0;}     // iteration: dp[i][j] = sum(dp[i-1][j - k*coins[i-1]])     // k = 0, 1, 2, ... , j / coins[i-1]    for (int i = 1; i <= coinKinds; ++i)    {         for (int j = 1; j <= sum; ++j)         {             dp[i][j] = 0;             for (int k = 0; k <= j / coins[i-1]; ++k)            {                dp[i][j] += dp[i-1][j - k * coins[i-1]];            }         }     }     return dp[coinKinds][sum]; }  int main() {    int coins[7] = {1, 2, 5, 10, 20, 50, 100};    int sum = 200;    int result = coinCombinations(coins, 7, 5);    cout << "using 8 kinds of coins construct 200, combinations are:  " << endl;    cout << result << endl;    return 0; }

扩展：

如果要输出所有的可能的组合结果？

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