题目链接：http://codeforces.com/problemset/problem/835/C
C. Star sky
time limit per test 2 seconds
memory limit per test 256 megabytes
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.
题目描述：
100*100的星空上有n颗星星，各自有一个初始亮度，且在0~C范围内周期性变化，给出q个查询，每个查询给出时间和一个矩形，求在该时间时在矩形内的星星的亮度和。
数据范围：
1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10
Solution:
由于数据范围较大，考虑预处理，使每次查询都能在O(1)的时间内得到矩形内的星星数量。
用sum[i][j][k]表示左上顶点为(1,1)，右下顶点为(i,j)的矩形范围内初始亮度为k的星星数量，转移方程：
sum[i][j][k] = sum[i][j][k] + (sum[i-1][j][k] + sum[i][j-1][k] - sum[i-1][j-1][k]);（也就是矩阵前缀和）
在每次查询时将初始亮度不同的星星分开处理，得到左下角坐标为(x,y)，右上角坐标为(xx,yy)构成的矩形内，初始亮度为k的星星数量的方程为：
ans = sum[xx][yy][k] + sum[x-1][y-1][k] - sum[x-1][yy][k] - sum[xx][y-1][k];
最后将每个t时刻星星的亮度乘以个数后再累加即可
代码：
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<algorithm>
#define mst(a, b) memset(a, b, sizeof(a))
#define rush() int T; scanf("%d", &T); while(T--)
using namespace std;
const int MaxN = 1e5 + 5;
const double eps = 1e-9;
int n, q, c;
int sum[105][105][15];
void pre_operation() {
for(int i = 1; i <= 100; i++)
for(int j = 1; j <= 100; j++)
for(int k = 0; k <= c; k++)
sum[i][j][k] += (sum[i-1][j][k] + sum[i][j-1][k] - sum[i-1][j-1][k]);
}
//得到该视野内初始亮度为k的星星的数量
int get_num(int k, int x, int y, int xx, int yy) {
int res = sum[xx][yy][k] + sum[x-1][y-1][k] - sum[x-1][yy][k] - sum[xx][y-1][k];
return res;
}
int main()
{
scanf("%d %d %d", &n, &q, &c);
for(int i = 1; i <= n; i++) {
int x, y, s;
scanf("%d %d %d", &x, &y, &s);
sum[x][y][s]++;
}
pre_operation(); //得到矩阵的前缀和sum数组
while(q--) {
int t, x, y, xx, yy;
scanf("%d %d %d %d %d", &t, &x, &y, &xx, &yy);
int ans = 0;
for(int i = 0; i <= c; i++) {
int tt = (i + t) % (c + 1);
ans += tt * get_num(i, x, y, xx, yy);
}
printf("%d\n", ans);
}
return 0;
}