动态规划：硬币凑值

硬币凑值

题目大意：给出n个硬币以及面值，求凑出面值为m所需要最少的硬币是多少

解题思路： 假设有面值1 3 5，从凑0个硬币开始想，d(0) ＝ 0；再想d(1) = d(1-1)+1 = 1；d(2) = d(2-1) + 1 = 2;
d(3) = min（d(3-1)+1，d(3-3)+1) = 1;
所以d(i) = min(d(i-y(x))+1) y(x)访问全部面值硬币

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int d[100];int n; //一共n种硬币int y[100]; //硬币面值int m; //要凑的面值int w(int x) {    int re = -1;    for(int i = 0; i < n; i++) {        if(d[x-y[i]] == -1 || x - y[i] < 0)            continue;        if(re == -1 || d[x-y[i]] + 1 < re)            re = d[x-y[i]] + 1;    }    return re;}int main() {    memset(d, 0, sizeof(d));    cin >> m;    cin >> n;    for(int i = 0; i < n; i++)        cin >> y[i];    for(int i = 1; i <= m; i++)        d[i] = w(i);    if(d[m] == -1)        cout << "凑不出";    else        cout << d[m];    return 0;}