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;
using ll = long long;
using ld = long double;
using db = double;
using str = string;
#define int ll
#define MN 1e5+5
#define MOD 1000000007
#define endl '\n'
#define endlfl endl << flush
#define dmp(x) cerr<<__LINE__<<" "<<#x<<" "<<x<<endl
const int INF=1e9 + 10;
const auto beg = std::chrono::high_resolution_clock::now();
void dbg_time() {
auto us = (double)std::chrono::duration_cast<std::chrono::microseconds>(std::chrono::high_resolution_clock::now() - beg).count();
cerr << "Operation time: " << (long double)us/1000000 << " seconds." << endl;
}
int n, k, q;
void solve()
{
cin >> n >> k;
vector<int> a(n+1);
vector<int> P(n+1, 0);
vector<int> Pmultl(n+1, 0);
vector<int> Pmultr(n+1, 0);
for (int i = 1; i <= n; ++i)
{
cin >> a[i];
P[i]=P[i-1]+a[i];
Pmultl[i]=Pmultl[i-1]+a[i]*(n-i+1);
Pmultr[i]=Pmultr[i-1]+a[i]*i;
}
cin >> q;
while(q--) {
int action; cin >> action;
if (action==1) {
int input; cin >> input;
} else {
int l, r, m; cin >> l >> r >> m;
int sum=(P[r]-P[l-1])*m;
sum-=(Pmultl[l+m-1]-Pmultl[l-1]) - (P[l+m-1]-P[l-1])*(n-(l+m-1)+1);
sum-=(Pmultr[r]-Pmultr[r-m+1]) - (P[r]-P[r-m+1])*(r-m+1);
cout << sum << endl;
}
}
return;
}
signed main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int tt=1;// cin >> tt;
while(tt--) solve();
dbg_time();
return 0;
}
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