Problem P09. [算法课动态规划] 换硬币

动态规划当前状态和之前状态息息相关。比如这题：组成硬币，已经组成好的硬币x加上面值为y的硬币就可以组成好x+y的硬币。所以要算出组成x+y的硬币，就要先算出组成x的硬币。可以用循环从x=0开始计算。

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

int nums[] = {2, 5, 7};
int x[105]={0};

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++){
        x[i]=-1;
    }
    int m = sizeof(nums)/4;
    for (int i = 1; i <= n; i++){
        for (int j = 0; j < m; j++){
            if (i-nums[j]>=0&&x[i-nums[j]]!=-1){
                int y = x[i-nums[j]]+1;
                if (x[i]==-1||x[i]>y){
                    x[i]=y;
                }
            }
        }
    }
    printf("%d", x[n]);
    return 0;
}