思路
类似完全背包,对于每种货币,枚举选几张。
#include <iostream>
using namespace std;
using LL = long long;
const int N = 30, M = 10010;
int n, m;
int w[N];
LL f[N][M];
int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) cin >> w[i];
f[0][0] = 1;
for (int i = 1; i <= n; i++)
for (int j = 0; j < M; j++)
for (int k = 0; k * w[i] <= j; k++)
f[i][j] += f[i - 1][j - k * w[i]];
cout << f[n][m] << endl;
return 0;
}