Submission #19092

#	Time	Username	Problem	Language	Result	Execution time	Memory
19092		cki86201	Evaluation (kriii1_E)	C++98	0 / 1	225 ms	57580 KiB

E

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>

using namespace std;
typedef long long ll;
typedef pair<int, int> Pi;
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define sz(x) (int)x.size()
#define rep(i,n) for(int i=0;i<n;i++)
#define all(x) x.begin(),x.end()

const int MOD = 1e9 + 7;

struct line{
	line(){}
	line(int x,int y1,int y2,int v):x(x), y1(y1), y2(y2), v(v){}
	int x, y1, y2, v;
	bool operator<(const line &l)const{
		return x < l.x;
	}
}L[200010];

struct node{
	node(){x = u = 0, left = 0;}
	int x, u;
	int left;
}T[4500030];
int Z;

int update(int rt, int s, int d, int l, int r, int v){
	if(l <= s && d <= r){
		T[rt].u = (T[rt].u + v + MOD) % MOD;
		int tmp = T[rt].x;
		T[rt].x = (T[rt].x + (ll)(d-s+1) * v) % MOD;
		if(T[rt].x < 0)T[rt].x += MOD;
		return tmp;
	}
	int m = (s+d)>>1, res = 0;
	if(!T[rt].left)T[rt].left = ++Z; ++Z;
	if(l<=m){
		res += update(T[rt].left, s, m, l, r, v);
	}
	if(m<r){
		res += update(T[rt].left + 1, m+1, d, l, r, v);
	}
	T[rt].x = (T[rt].x + (ll)(min(d, r) - max(s, l) + 1) * v) % MOD;
	if(T[rt].x < 0)T[rt].x += MOD;
	res = (res + (ll)(min(d, r) - max(s, l) + 1) * T[rt].u) % MOD;
	return res;
}

int main(){
	int n; scanf("%d", &n);
	rep(i, n){
		int a, b, c, d, p;
		scanf("%d%d%d%d%d", &a, &b, &c, &d, &p);
		L[i+i] = line(a, b, d, p);
		L[i+i+1] = line(c+1, b, d, -p);
	}
	sort(L, L+n+n);
	int root = 0;
	ll ans = 0;
	ll now = 0;
	for(int i=0;i<n+n;i++){
		if(i != 0){
			ans += now * (L[i].x - L[i-1].x), ans %= MOD;
		}
		now += (ll)(L[i].y2 - L[i].y1 + 1) * L[i].v * L[i].v;
		now += 2LL * L[i].v * update(root, 1, 1e9, L[i].y1, L[i].y2, L[i].v);
		now %= MOD;
		if(now < 0)now += MOD;
	}
	printf("%lld", ans);
	return 0;
}

• Subtask 1: 0 / 1