CodeForces - 835C Star sky （前缀和）

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

C. Star sky

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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;;
	}
}