Codeforces Round #427 (Div. 2) C. Star sky(前缀和）-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_34826781/article/details/76612964

本文介绍了一种模拟星空亮度的方法，利用前缀和思想处理星星亮度随时间变化的问题，并通过三维数组记录不同时间点星星亮度，最终实现高效计算指定矩形区域内的总亮度。

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The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.

You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.

A star lies in a rectangle if it lies on its border or lies strictly inside it.

Input

The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.

The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si (1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.

The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.

Output

For each view print the total brightness of the viewed stars.

Examples

input
2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5

output
3
0
3

input
3 4 5
1 1 2
2 3 0
3 3 1
0 1 1 100 100
1 2 2 4 4
2 2 1 4 7
1 50 50 51 51

output
3
3
5
0

Note

Let's consider the first example.

At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.

At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.

At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.

题目大意：在一个二维坐标中存在着一些星星，星星的闪亮度每隔一秒增加1，到达c后变为零重新增加，有一个人在某个时刻观察矩形范围内的天空，要求输出这个矩形内的闪亮度

解题思路：这道题的数据导致没有办法通过暴力枚举的方式解决，所以只能使用前缀和的思想去解决，考虑到在不同时刻的闪亮度不同，所以要建造一个三维的数组，然后求出前缀和，通过给的观察的矩形求出

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long LL;

int n,q,c,i,t,xx,yy,x,y,j,k;
int tu[15][105][105],a[15][105][105];
struct point
{
	int x,y,k;
}star[100005];
int main()
{
	cin>>n>>q>>c;
	for(i=1;i<=n;i++)
	{
		cin>>star[i].x>>star[i].y>>star[i].k;
		tu[0][star[i].x][star[i].y]+=star[i].k;
	}
	for(i=1;i<=c;i++)
	{
		for(j=1;j<=n;j++)
		{
			star[j].k++;
			tu[i][star[j].x][star[j].y]+=star[j].k%(c+1);
		}
	}
	for(i=0;i<=c;i++)
	{
		for(j=1;j<=100;j++)
		{
			for(k=1;k<=100;k++)
			{
				a[i][j][k]+=tu[i][j][k]+a[i][j][k-1];
			}
		}
		for(j=1;j<=100;j++)
		{
			for(k=1;k<=100;k++)
			{
				a[i][j][k]+=a[i][j-1][k];
			}
		}
	}
	while(q--)
	{
		cin>>t>>x>>y>>xx>>yy;
		t%=(c+1);
		cout<<a[t][xx][yy]-a[t][x-1][yy]-a[t][xx][y-1]+a[t][x-1][y-1]<<endl;
	}
	return 0;
}