codeforces 835C Star sky(二维树状数组)

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.

比较裸的二维树状数组，有一个思维点在于需要取模。每次查询的值不一样，但是因为c最多只有10，所以记录下0-10分别是哪个值即可。

#include <algorithm>
#include <cstdio>
#include <cmath>
#include <map>
using namespace std;

const int MAXN=110;
int a[11][MAXN][MAXN];
int lowbit(int i){
return i&-i;
}
int c;
int nn=100;
int t;
int tt;
void add(int x,int y,int num){
for(int i=x;i<=nn;i+=lowbit(i)){
for(int j=y;j<=nn;j+=lowbit(j)){
a[tt][i][j]+=(num+tt)%(c+1);
}
}
}

int sum(int i, int j){
int result = 0;
for(int x = i; x > 0; x -= lowbit(x)) {
for(int y = j; y > 0; y -= lowbit(y)) {
result += a[tt][x][y];
}
}
//printf("%d %d %d\n",i,j,result);
return result;
}
int main(){
int n,q;
scanf("%d%d%d",&n,&q,&c);
for(int i=0;i<n;i++){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
for(tt=0;tt<=c;tt++){
add(x,y,z);
}
}
for(int i=0;i<q;i++){
int x1,y1,x2,y2;
scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);
tt=t%(c+1);
printf("%d\n",sum(x2,y2)+sum(x1-1,y1-1)-sum(x1-1,y2)-sum(x2,y1-1));
}
}