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.
題意：給出n個星星的位置和初始亮度，詢問一個矩形內，星星的亮度和是多少（亮度隨時間增長，到最大值mx後，下一秒變成0）。
題解：二維前綴和。
代碼：
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#include<math.h>
using namespace std;
typedef long long int ll;
typedef pair<int,int>pa;
const int N=1e5+10;
const int mod=1e9+7;
const ll INF=1e18;
int read()
{
int x=0;
char ch = getchar();
while('0'>ch||ch>'9')ch=getchar();
while('0'<=ch&&ch<='9')
{
x=(x<<3)+(x<<1)+ch-'0';
ch=getchar();
}
return x;
}
/***********************************************************/
int p[12][110][110];
int n,q,c,x,y,z;
int t;
int x2,y2;
int main()
{
memset(p,0,sizeof(p));
scanf("%d%d%d",&n,&q,&c);
for(int i=1; i<=n; i++)
{
scanf("%d%d%d",&x,&y,&z);
p[z][x][y]++;
}
for(int k=0; k<=10; k++)
{
for(int i=1; i<=100; i++)
{
for(int j=2; j<=100; j++)
{
p[k][i][j]+=p[k][i][j-1];
}
}
}
for(int k=0; k<=10; k++)
{
for(int j=1; j<=100; j++)
{
for(int i=2; i<=100; i++)
{
p[k][i][j]+=p[k][i-1][j];
}
}
}
while(q--)
{
scanf("%d%d%d%d%d",&t,&x,&y,&x2,&y2);
int ans,op;
op=0;
for(int k=0; k<=10; k++)
{
ans=0;
ans+=p[k][x2][y2];
ans+=p[k][x-1][y-1];
ans-=p[k][x2][y-1];
ans-=p[k][x-1][y2];
op+=((k+t)%(c+1))*ans;
}
printf("%d\n",op);
}
return 0;
}