AcWing 281. 硬币

题目链接：​​传送门​​

只是询问一个可行性
那么二进制拆分加一个bitset就行了

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

using namespace std;
typedef long long ll;
int n, m, a[A], c; bitset<A> f;

int main(int argc, char const *argv[]) {
  while (cin >> n >> m and n and m) {
    int ans = 0; f.reset(); f[0] = 1;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) {
      cin >> c;
      for (int k = 1; c; k <<= 1) {
        if (k > c) k = c;
        c -= k;
        f |= (f << (k * a[i]));
      }
    }
    for (int i = 1; i <= m; i++) if (f[i]) ans++;
    cout << ans << endl;
  }
}