Codeforces835C Star sky_mdyd835-优快云博客

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.

————————————————————————————————————

题目的意思是有一片100*100的星空，上面有n颗星星，每个星星有一个亮度，且在

0~C范围内周期性变化，现在给出q个查询，每个查询给出时间和一个矩形，求在该

时间时矩形内星星的亮度和。

思路：先预处理处每个区域每种亮度的星星有多少颗，可以直接搞波前缀和，然后

枚举各个亮度在查询区域内的星星个数，计算总和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
#define mod 10000007
#define mem(a,b) memset(a,b,sizeof a)

int main()
{
int n,q,c,x,y,x1,x2,y1,y2,c1,t;
int mp[105][105][15];

while(~scanf("%d%d%d",&n,&q,&c))
{
    mem(mp,0);

    for(int i=0;i<n;i++)
    {
        scanf("%d%d%d",&x,&y,&c1);
        mp[x][y][c1]++;
    }
    for(int i=0;i<=10;i++)
    {
        for(int j=1;j<102;j++)
        {
            for(int k=1;k<102;k++)
            {
                mp[j][k][i]+=(mp[j-1][k][i]+mp[j][k-1][i]-mp[j-1][k-1][i]);
            }
        }
    }
    while(q--)
    {
        scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);
        int ans=0;
        for(int i=0;i<=10;i++)
        {
            int cnt=mp[x2][y2][i]-mp[x1-1][y2][i]-mp[x2][y1-1][i]+mp[x1-1][y1-1][i];
            int xx=(t+i)%(c+1);
            ans+=(cnt*xx);
        }
        printf("%d\n",ans);
    }


}

    return 0;
}