CodeForces - 835C - Star sky （dp + 周期）

• The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (x**i, y**i), a maximum brightness c, equal for all stars, and an initial brightness s**i (0 ≤ s**i ≤ c).

  Over time the stars twinkle. At moment 0 the i-th star has brightness s**i. 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 t**iand 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 x**i, y**i, s**i (1 ≤ x**i, y**i ≤ 100, 0 ≤ s**i ≤ 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 t**i, x1i, y1i, x2i, y2i (0 ≤ t**i ≤ 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.

Example

• 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.

  \

  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;
  }
  

  
  \

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