제출 #1254413

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1254413		onur8ocak	Addk (eJOI21_addk)	C++20	0 / 100	71 ms	2888 KiB

Main

#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
#define int long long
#define F first
#define S second 
#define pb push_back
#define endl '\n'
#define all(v) v.begin(),v.end()
#define gcd(a,b) __gcd(a,b)
#define mt make_tuple
#define pqueue priority_queue
typedef pair<int,int> ii;
typedef tuple<int,int,int> iii;
typedef tuple<int,int,int,int> iiii;
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef vector<char> vc;
typedef vector<iii> viii;
typedef set<int> si;
typedef vector<ii> vii;				
typedef vector<vi> vvi;
typedef vector<si> vsi;
typedef vector<vb> vvb;
typedef vector<vc> vvc;


int32_t main(){
	int n,s;
	cin >> n >> s;
	vi a(n+10);
	for(int i=1;i<=n;i++) cin >> a[i];
	vi pre(n+10,0),suf(n+10,0),pre2(n+10,0),suf2(n+10,0);
	for(int i=1;i<=n;i++) pre[i]=pre[i-1]+a[i];
	for(int i=n;i>=1;i--) suf[i]=suf[i+1]+a[i];
	for(int i=1;i<=n;i++){ 
		int val=a[i];
		pre2[i]=pre2[i-1]+(i*val);
	}
	for(int i=n;i>=1;i--){
		int val=a[i];
		suf2[i]=suf2[i+1]+((n-i+1)*val);
	}
	int q;
	cin >> q;
	while(q--){
		int k,l,r,mid;
		cin >> k >> l >> r >> mid;
		if(k==1) continue;
		int fark=(r-l)+1;
		if(mid==1||mid==fark){
			cout << pre[r]-pre[l-1] << endl;
			continue;
		}
		int l1,r1,l2,r2,m1,m2,sum=0;
		if(fark<2*mid-1) mid=(fark+1)-mid;
		//cout << mid << endl;
		l1=l;
		r1=l1+(mid-2);
		m1=r1+1;
		r2=r;
		l2=r2-(mid-2);
		m2=l2-1;
		int len = fark - 2 * (mid - 1);
		/*if (m1 <= m2 && m2 - m1 + 1 >= len)
		sum += mid * (pre[m1 + len - 1] - pre[m1 - 1]);*/
		int sonuc=pre2[r1]-pre2[l1-1];
		int tmp=pre[r1]-pre[l1-1];
		tmp*=(l1-1);
		sum+=(sonuc-tmp);
		//cout << (sonuc-tmp) << endl;
		sum+=(mid*(pre[m2]-pre[m1-1]));
		//cout << (mid*(pre[m2]-pre[m1-1]));
		sonuc=suf2[l2]-suf2[r2+1];
		tmp=suf[l2]-suf[r2+1];
		tmp*=(n-r2);
		sum+=(sonuc-tmp);
		//cout << (sonuc-tmp) << endl;
		cout << sum << endl;
	}
	return 0;
}

• Subtask 1: 0 / 36
• Subtask 2: 0 / 56
• Subtask 3: 0 / 8