codeforces 835 C Star sky

C. Star sky

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 31 1 13 2 02 1 1 2 20 2 1 4 55 1 1 5 5

output

303

input

3 4 51 1 22 3 03 3 10 1 1 100 1001 2 2 4 42 2 1 4 71 50 50 51 51

output

3350

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：http://codeforces.com/contest/835/problem/C 
大致题意： 

给出n个star的位置(xi,yi)和初始亮度si，以及最大亮度c。star的亮度单位时间增长1，达到最大亮度之后归0。q个查询，每个查询包含一个矩阵和时间t，求矩阵的亮度之和。
解法： 
先做一个预处理，求出sum[x][y][t]，表示矩形(0,0)~(x,y) 在时间t时的所有点的亮度总和，sum[x][y][t] += sum[x-1][y][t] + sum[x][y-1][t] - sum[x-1][y-1][t]。

之后对于每个查询就可以在O(1)时间内求出，ans = sum[rx][ry][t] - sum[rx][ly-1][t] - sum[lx-1][ry][t] + sum[lx-1][ly-1][t]

代码：

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;const int N = 1e5 + 10;using P = pair<int,int>;P p[N];int s[N];int n,q,c;long long sum[105][105][15]{0};void add(long long *,long long * , int);void sub(long long *,long long * , int);void init();int main(){    scanf("%d%d%d",&n,&q,&c);    for(int i = 0 ; i < n; i++){        scanf("%d%d%d",&p[i].first,&p[i].second,&s[i]);    }    init();    int t,lx,ly,rx,ry;    for(int i = 0 ; i < q ; i++){        scanf("%d%d%d%d%d",&t,&lx,&ly,&rx,&ry);        t %= (c+1);        lx--;ly--;        long long ans = sum[rx][ry][t] - sum[rx][ly][t] - sum[lx][ry][t] + sum[lx][ly][t];        printf("%I64d\n",ans);    }}void init(){    for(int i = 0 ; i < n ; i++){        for(int j = 0 ; j <=c ; j++){            sum[p[i].first][p[i].second][j] += (s[i] + j) % (c+1);        }    }    for(int i = 1 ; i <= 100 ; i++){        for(int j = 1 ; j <= 100 ; j++){            if(i == 1 && j == 1) continue;            else if(i == 1) add(sum[i][j],sum[i][j-1],c+1);            else if(j == 1) add(sum[i][j],sum[i-1][j],c+1);            else{                add(sum[i][j],sum[i][j-1],c+1);                add(sum[i][j],sum[i-1][j],c+1);                sub(sum[i][j],sum[i-1][j-1],c+1);            }        }    }}void add(long long *s1 ,long long *s2 ,int len){    for(int i = 0 ; i < len ; i++){        s1[i]+=s2[i];    }}void sub(long long *s1, long long *s2, int len){    for(int i = 0 ; i < len ; i++){        s1[i]-=s2[i];    }}