Codeforces Round #427 (Div. 2) C. Star sky（前缀和）

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.
题意：给出n个星星的位置和初始亮度，询问一个矩形内，星星的亮度和是多少（亮度随时间增长，到最大值mx后，下一秒变成0）。
题解：二维前缀和。
代码：

#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;}