solution

DP。箱は大きい方から貪欲に使って$\mathrm{dp}: (M+1) \times 2^N \to 2$。$O(M4^N)$。

implementation

#include <iostream>
#include <vector>
#include <algorithm>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
#define repeat_from(i,m,n) for (int i = (m); (i) < (n); ++(i))
using namespace std;
template <typename T, typename X> auto vectors(T a, X x) { return vector<T>(x, a); }
template <typename T, typename X, typename Y, typename... Zs> auto vectors(T a, X x, Y y, Zs... zs) { auto cont = vectors(a, y, zs...); return vector<decltype(cont)>(x, cont); }
int main() {
// input
int n; cin >> n;
vector<int> a(n); repeat (i,n) cin >> a[i];
int m; cin >> m;
vector<int> b(m); repeat (j,m) cin >> b[j];
// compute
vector<int> sum(1<<n); repeat_from (s,1,1<<n) sum[s] = sum[s&(s-1)] + a[__builtin_ctz(s)]; // http://www.slideshare.net/KMC_JP/slide-www
sort(b.rbegin(), b.rend());
vector<vector<bool> > dp = vectors(bool(), m+1, 1<<n);
dp[0][0] = true;
repeat (j,m) {
repeat (k,1<<n) if (sum[k] <= b[j]) {
repeat (s,1<<n) {
if (dp[j][s]) dp[j+1][s|k] = true;
}
}
}
// output
int ans = -1;
repeat (j,m+1) {
if (dp[j][(1<<n)-1]) {
ans = j;
break;
}
}
cout << ans << endl;
return 0;
}