CodeForces - 835C Star sky （前缀和）

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这是一道关于计算星星在不同时间点矩形区域内总亮度的问题，涉及动态变化的亮度和坐标轴上的矩形覆盖。题目要求在给定的最大亮度、星星数量和观察次数下，计算每个观察时刻矩形内的星星总亮度。解决方法是通过记录不同亮度的星星个数前缀和，根据时间关系计算亮度，并注意亮度超过最大值后会归零的特性。

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题目链接：http://codeforces.com/problemset/problem/835/C点击打开链接

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.

这道题跟poj 2352 && hdu 1541 Stars很像就是多了亮度因此多开一维

可以用树状数组来记录前缀和但是数据量比较小可以不用树状数组也忘了。。

一开始的时候第三维我记录的是时间也就是在某一时刻整个星空星星亮度情况的前缀和

但是wa了后面才发现可能存在重复星星在同一个点上这样一来便无法处理

因此将第三维换做记录亮度处理星星个数的前缀和

然后通过关系计算即可

注意因为题目中是到了c+1才变为0

因此是%c+1

#include <bits/stdc++.h>
using namespace std;
int a[111][111][11];
int main()
{
	for(int i=0;i<111;i++)
		for(int j=0;j<111;j++)
			for(int k=0;k<11;k++)
			a[i][j][k]=0;
	int n,m,c;
	cin >> n>>m >>c;
	for(int i=0;i<n;i++)
	{
		int x,y,b;
		cin >> x >> y >>b;
		a[x][y][b]++;
	}
	for(int k=0;k<c+1;k++)
	{
		for(int i=1;i<111;i++)
		{
			for(int j=1;j<111;j++)
			{
				a[i][j][k]+=a[i][j-1][k]+a[i-1][j][k]-a[i-1][j-1][k];
			}
		}
	}
	for(int i=0;i<m;i++)
	{
		int mid[5];
		cin >> mid[0] >> mid[1] >> mid[3]>>mid[2] >> mid[4];
		int ans=0;
		for(int i=0;i<=c;i++)
		{
			ans+=((i+mid[0])%(c+1))*(a[mid[2]][mid[4]][i]-a[mid[1]-1][mid[4]][i]-a[mid[2]][mid[3]-1][i]+a[mid[1]-1][mid[3]-1][i]);
		}
		cout << ans <<endl;;
	}
}