C. Star sky Codeforces

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 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.

题意：在坐标中给了你很多带有亮度的星星。这些星星随着时间做周期性变化。然后给出q个询问，每一个询问包含了查询区域的位置与当前时间，让你求出这段区域的总亮度是什么。

思路：

预处理到每一个从原点出发到每一个点所围的矩形内星星的个数。

dp【x】【y】【z】就是从1,1点出去到x，y所形成的矩阵中（包括边界）的亮度为z的星星的个数

所以dp【X】【Y】【Z】=dp【X】【Y】【Z】+dp【X-1】【Y】【C】+dp【X】【Y-1】【C】-DP【X-1】【Y-1】【C】

#include<bits/stdc++.h>
using namespace std;
int dp[120][120][11];
int n,q,c;
void init()
{
for(int i=1; i<=100; i++)
for(int j=1; j<=100; j++)
for(int c=0; c<=10; c++)
dp[i][j][c]+=dp[i-1][j][c]+dp[i][j-1][c]-dp[i-1][j-1][c];
}
int main()
{

while(scanf("%d%d%d",&n,&q,&c)!=EOF)
{
memset(dp,0,sizeof(dp));
for(int i=0; i<n; i++)
{
int x,y,b;
scanf("%d%d%d",&x,&y,&b);
dp[x][y][b]++;
}
init();
for(int j=0; j<q; j++)
{
int x,y,x1,y1,t;
int ans=0;
int sum=0;
scanf("%d%d%d%d%d",&t,&x,&y,&x1,&y1);
for(int i=0; i<=c; i++)
{
ans=dp[x1][y1][i]+dp[x-1][y-1][i]-dp[x-1][y1][i]-dp[x1][y-1][i];
ans*=(t+i)%(c+1);
sum+=ans;
}
printf("%d\n",sum);

}

}
}