C. Star sky【前缀和预处理】

题目点这里
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.
题意：
在一个二维平面内有n个星星，每个星星有一个亮度值，在初始时为s，每过一秒数值增加1，数值不超过c，当超过c时亮度值变为0，然后再次循环。询问q次，每次询问在t时刻从左下角（x1,y1）到右上角(x2,y2)的矩形区域内所有星星的亮度总和是多少。
思路：
坐标和c值都比较小，可以枚举c预处理降低时间复杂度，利用了简单的容斥思想。枚举c，计算c秒下从原点到（x, y）的亮度总和，最后减一下就可以了。题目不难，坑点就是有重合点。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define max_n 100010using namespace std;typedef long long LL;int dp[15][110][110], mapp[15][110][110];int main() {    int n, q, c, x, y, z;    memset(dp, 0, sizeof(dp));    scanf("%d %d %d", &n, &q, &c);    for(int i = 0; i < n; i++) {        scanf("%d %d %d", &x, &y, &z);        for(int p = 0; p <= c; p++) {            mapp[p][x][y] += (z + p) % (c + 1); //有重点，记得去重         }    }    for(int p = 0; p <= c; p++) { //预处理，优化时间         for(int i = 1; i <= 100; i++) {            for(int j = 1; j <= 100; j++) {                dp[p][i][j] = mapp[p][i][j] + dp[p][i - 1][j] + dp[p][i][j - 1] - dp[p][i - 1][j - 1]; //容斥的思想             }        }    }    while(q--) {        int t, x1, x2, y1, y2;        scanf("%d %d %d %d %d", &t, &x1, &y1, &x2, &y2);        t %= (c + 1);        int sum = dp[t][x2][y2] - dp[t][x1 - 1][y2] - dp[t][x2][y1 - 1] + dp[t][x1 - 1][y1 - 1];        printf("%d\n", sum);    }    return 0;}