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"
using namespace std;
#define rep(i, a, b) for(int i=a; i<=b; i++)
#define trav(a, x) for(auto& a : x)
#define all(x) begin(x), end(x)
#define sz(x) (int) x.size()
#define pb push_back
#define f first
#define s second
#define nl "\n"
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
const int MOD = 1e9+7;
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;
// dp[i][mask] := is it possible to cover first i requests with b_mask
// if leftover > 0- check dp[i][mask^(1<<j)] for all j in subset
// if leftover=0- check dp[i-1][mask^(1<<j)]
// too slow..
// dp[mask]=max prefix length that can be covered WITH LEFTOVER < next a_i
int n, m;
int a[20];
int b[20];
int ps[20];
int sum[1048576];
int dp[1048576];
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
cin >> n >> m;
rep(i, 0, n-1) cin >> a[i];
rep(j, 0, m-1) cin >> b[j];
rep(i, 0, n-1){
ps[i]=(i?ps[i-1]:0)+a[i];
}
rep(i, 1, (1<<m)-1){
rep(j, 0, m-1){
if(i&(1<<j)){
if(sum[i]==0) sum[i]=sum[i^(1<<j)]+b[j];
int x=dp[i^(1<<j)];
int r=sum[i^(1<<j)]-(x==0?0:ps[x-1]);
if(r+b[j]>a[x]) continue;
if(r+b[j]==a[x]) dp[i]=max(dp[i], x+1);
else dp[i]=max(dp[i], x);
}
}
if(dp[i]==n){
cout << "YES"; return 0;
}
}
cout << "NO";
}
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