이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define fi first
#define se second
#define ll long long
#define ld long double
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define mpp make_pair
#define ve vector
using namespace std;
using namespace __gnu_pbds;
template<class T> using oset = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
const ll inf = 1e18; const int iinf = 1e9;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T> inline bool chmin(T& a, T b) { return (a > b ? a = b, 1 : 0); }
template <typename T> inline bool chmax(T& a, T b) { return (a < b ? a = b, 1 : 0); }
inline void solve() {
int n, m;
cin >> n >> m;
ve<int> a(n), b(m);
for (auto &i : a) cin >> i;
for (auto &i : b) cin >> i;
ve<ve<int>> dp(n + 1, ve<int> (1 << m));
dp[0][0] = 1;
ve<int> sm(1 << m);
ve<ve<int>> sums(1001);
for (int ma = 1; ma < (1 << m); ++ma) {
sm[ma] = sm[ma - (ma & -ma)] + b[__builtin_ctz(ma)];
if (sm[ma] > 1000) continue;
sums[sm[ma]].pb(sm[ma]);
}
for (int i = 0; i < n; ++i) {
for (int ma = 0; ma < (1 << m); ++ma) {
if (!dp[i][ma]) continue;
for (int &f : sums[a[i]]) {
if ((ma & f) == 0) {
dp[i + 1][ma | f] = 1;
}
}
}
}
bool ok = 0;
for (int i = 0; i < (1 << m); ++i) {
ok |= dp[n][i];
}
cout << (ok ? "YES" : "NO");
}
signed main() {
ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int q = 1; // cin >> q;
while (q--) solve();
cerr << fixed << setprecision(3) << "Time execution: " << (double)clock() / CLOCKS_PER_SEC << endl;
}
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