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;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T> using oset =
tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define ll long long
#define ld long double
#define ar array
#define vi vector<int>
#define vii vector<vector<int>>
#define pii pair<int, int>
#define pb push_back
#define all(x) x.begin(), x.end()
#define f first
#define s second
const int MOD = 1e9+7;
const int inf = 1e9;
const ll linf = 1e18;
const int d4i[4]={-1, 0, 1, 0}, d4j[4]={0, 1, 0, -1};
const int d8i[8]={-1, -1, 0, 1, 1, 1, 0, -1}, d8j[8]={0, 1, 1, 1, 0, -1, -1, -1};
// -------------------------------------------------- Main Code --------------------------------------------------
int a[20], b[20];
int n, m;
pii sol(int mask) {
if (mask == 0) return {0, 0};
pii ans = {0, 0};
for (int i = 0; i < m; i++) {
if (mask&(1<<i)) {
pii x = sol(mask^(1<<i));
int prevPer = x.f;
int left = x.s;
if (left+b[i] == a[prevPer]) prevPer++, left = 0;
else if (left+b[i] < a[prevPer]) left += b[i];
ans = max(ans, {prevPer, left});
}
}
return ans;
}
void sol() {
cin >> n >> m;
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = 0; i < m; i++) cin >> b[i];
for (int i = 0; i < (1<<m); i++) {
if (sol(i).f == n) {
cout << "YES" << endl;
return;
}
}
cout << "NO" << endl;
}
int main () {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
// cin >> t;
while (t--) {
sol();
}
return 0;
}
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