Codeforces Round #427 (Div. 2) C

C. Star sky
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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.

#include <bits/stdc++.h>using namespace std;int a[125][125][11];int main(){    int n,q,c;    scanf("%d %d %d",&n,&q,&c);    while(n--)    {        int x,y,t;        scanf("%d %d %d",&x,&y,&t);        int h=t;        for(int i=0;i<=(c);i++)        {            if(h>c)            {                h=0;            }            a[x][y][i]+=h;            h++;        }    }    for(int i=1;i<=100;i++)    {        for(int j=1;j<=100;j++)        {            for(int k=0;k<=(c);k++)            {                a[i][j][k]+=a[i][j-1][k];            }        }    }    while(q--)    {        int t,x1,y1,x2,y2;        int z=0;        scanf("%d %d %d %d %d",&t,&x1,&y1,&x2,&y2);        for(int i=x1;i<=x2;i++)        {            z+=(a[i][y2][t%(c+1)]-a[i][y1-1][t%(c+1)]);        }        printf("%d\n",z);    }    return 0;}