Star sky CodeForces - 835C （二维前缀和）

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最新推荐文章于 2021-05-24 15:38:04 发布

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分类专栏： # 前缀和

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本文介绍了一个星空模拟问题的解决方法，通过三维数组记录星星的位置和亮度变化，实现快速查询任意时刻指定矩形区域内的总亮度。

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Star sky

CodeForces - 835C

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_i and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x_1i, y_1i) and the upper right — (x_2i, y_2i). 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, x_1i, y_1i, x_2i, y_2i (0 ≤ t_i ≤ 10⁹, 1 ≤ x_1i < x_2i ≤ 100, 1 ≤ y_1i < y_2i ≤ 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

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

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：

给出 n个星星的坐标和它们的亮度si，星星最大亮度为c，t秒后星星的亮度为（si+t）%（c+1），q次询问，每次询问一个矩形区域内t秒后星星的亮度和。

思路：

开一个三维数组，记录每个前缀亮度为si的星星的个数即可

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int sum[105][105][12];
int main(){
    int n,q,c;
    while(cin >> n >> q >> c){
        memset(sum,0,sizeof(sum));
        for(int i = 1; i <= n; i++){
            int x,y,v;
            cin >> x >> y >> v;
            sum[x][y][v]++;
        }
        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];
                }
            }
        }
        while(q--){
            int t,sx,sy,ex,ey,ans = 0;
            cin >> t >> sx >> sy >> ex >> ey;
            for(int i = 0; i <= c; i++){
                int x = (i + t) % (c + 1);
                ans += x * (sum[ex][ey][i] - sum[sx-1][ey][i] - sum[ex][sy-1][i] + sum[sx-1][sy-1][i]);
            }
            cout << ans << endl;
        }
    }
    return 0;
}