找零钱

时限：1000ms 内存限制：10000K  总时限：3000ms

描述：

Poor Bessie has taken a job in the convenience store located just over the border in Slobbovia. Slobbovians use different coinages than the USA; their coin values change day-by-day!
Help Bessie make optimal change for Slobbovian shoppers. You will need to create C (1 <= C <= 1000) cents of change using N (1 <= N <= 10) coins of various values. All test cases will be solvable using the supplied coins.
If 5 coins of values 50, 25, 10, 5, and 1 were available, Bessie would make optimum change (minimal coins) of 93 cents by using 1 x 50, 1 x 25, 1 x 10, 1 x 5, and 3 x 1 coins (a total of 7 coins).
How hard could it be? The final two test cases will be challenging.

输入：

Line 1: Two space-separate integers: C and N
Lines 2..N+1: Each line contains a single unique integer that is a coin value that can be used to create change

输出：

Line 1: A single integer that is the minimum number of coins to create C cents

输入样例：

93 5
25
50
10
1
5

输出样例：

7
#include <iostream>#include <climits>using namespace std;void MinCoins(int coins[], int n, int money){int *sumTo = new int[money + 1];sumTo[0] = 0;for (int i = 1; i <= money; i++){sumTo[i] = INT_MAX;for (int j = 0; j < n; j++){if (i >= coins[j] && sumTo[i - coins[j]] + 1 < sumTo[i])sumTo[i] = sumTo[i - coins[j]] + 1;}}cout << sumTo[money] << endl;delete []sumTo;}int main(int argc, char *argv[]){int money, n;cin >> money >> n;int *coins = new int[n];for (int i = 0; i < n; i++)cin >> coins[i];MinCoins(coins, n, money);delete []coins;return 0;}