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 A 100010 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; } }