이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("Ofast,unroll-loops,fast-math,O3")
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <cassert>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <fstream>
#include <unordered_map>
#define ll long long
#define int long long
#define all(v) v.begin(), v.end()
#define nl '\n'
#define pb push_back
#define sz(s) (int)(s).size()
#define f first
#define s second
using namespace std;
const ll N = 1e6 + 50, MX = 1e18;
void solve(){
ll n;
cin >> n;
ll a[n + 1], b[n + 1];
for(int i = 1; i <= n; i++){
cin >> a[i] >> b[i];
}
ll ans = 0;
for(ll mask = 0; mask < (1 << n); mask++){
ll mx=0, mn=MX;
ll sum = 0;
for(int i = 0; i < n; i++){
if((mask >> i)&1){
mx = max(mx, a[i+1]);
mn = min(mn, a[i+1]);
sum += b[i+1];
}
}
if(mn == MX){
continue;
}
ans = max(sum - (mx - mn), ans);
}
cout << ans << nl;
}
signed main(){
//freopen("lca.in", "r", stdin);
//freopen("lca.out", "w", stdout);
ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
ll ql=1;
//cin >> ql;
//tst++;
while(ql--){
solve();
}
}
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