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 tiand 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.
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.
For each view print the total brightness of the viewed stars.
2 3 3 1 1 1 3 2 0 2 1 1 2 2 0 2 1 4 5 5 1 1 5 5
3 0 3
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
3 3 5 0
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.
注意星星到达最大亮度之后就会变回0；
开三维数组第一个变量代表时间。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int ma[12][110][110];
int n ,q ,c;
int main(){scanf("%d%d%d", &n, &q, &c);int x, y, s;for(int i = 0; i < n; i++){scanf("%d%d%d", &x,&y ,&s);for(int k = 0; k <= 10; k++){ //初始化11个时间 ma[k][x][y] += (s + k) % ( c + 1); //应该是+= }}for(int k = 0; k <= 10; k++)for(int i = 1; i <= 100; i++)for(int a = 1; a <= 100; a++){ma[k][i][a] += ma[k][i][a-1] + ma[k][i-1][a] - ma[k][i-1][a-1]; //写错加等于自己 } int x1, y1, x2, y2, ti;while(q--){scanf("%d%d%d%d%d", &ti, &x1, &y1, &x2, &y2);ti %= (c + 1);int te = ma[ti][x2][y2] - ma[ti][x1 -1][y2] -ma[ti][x2][y1-1] + ma[ti][x1 -1][y1 - 1];printf("%d\n", te);}return 0;
}