이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//#pragma GCC optimize("Ofast,O3,unroll-loops")
//#pragma comment(linker, "/SET: 2000000")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define bp __builtin_popcountll
#define mp make_pair
#define pb push_back
#define vec vector
#define ll long long
#define F first
#define S second
#define ld long double
#define pii pair <int, int>
#define elif else if
#define sz(a) int32_t(a.size())
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define fl fflush(stdout)
#define ex exit(0)
#define pll pair <ll, ll>
#define int ll
#define FOR(i, l, r) for(int i = l; i <= r; ++i)
#define ROF(i, l, r) for(int i = r; i >= l; --i)
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define ti tuple<int, int, int>
//#define FELIX
using namespace std;
using namespace __gnu_pbds;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rnd(228355);
const int N = 21;
const int M = (1 << N);
int d[M], a[N], b[N], n, m;
void solve(){
cin >> n >> m;
FOR(i, 0, n - 1) cin >> a[i];
FOR(i, 0, m - 1) cin >> b[i];
FOR(i, 1, n - 1){
a[i] += a[i - 1];
}
a[n] = 1e9;
int mx = -1;
int cur = (1 << m) - 1;
//cout << "CUR: " << cur << "\n";
FOR(i, 0, cur){
int sum = 0;
FOR(j, 0, m - 1){
int x = (1 << j);
if((i & x)) sum += b[j];
}
FOR(j, 0, m - 1){
int x = (1 << j);
if(!(i & x)){
if(sum + b[j] == a[d[i]]){
d[i ^ x] = max(d[i ^ x], d[i] + 1);
}
if(sum + b[j] < a[d[i]]){
d[i ^ x] = max(d[i ^ x], d[i]);
}
}
}
mx = max(mx, d[i]);
}
if(mx == n) cout << "YES";
else cout << "NO";
}
signed main(){
#ifdef FELIX
auto _clock_start = chrono::high_resolution_clock::now();
#endif
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t = 1;
//cin >> t;
while(t--) solve();
#ifdef FELIX
cerr << "Executed in " << chrono::duration_cast<chrono::milliseconds>(
chrono::high_resolution_clock::now()
- _clock_start).count() << "ms." << endl;
#endif
}
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