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.
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.
For each view print the total brightness of the viewed stars.
2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5
3
0
3
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
3
3
5
0
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：给你一个范围为100*100的天空，其中有很多星星，输入：n行为星星的亮度值，m行为查询时间t时区间（x1，y1) （x2，y2）的矩形范围内的星星的亮度值总和
题解：用一个三维数组sum[i][i][k]来表示在坐标（i，j）内的亮度为k的星星的数量，相当于二维前缀和表示了到i，j这个坐标的星星总数，再根据二维前缀和的容斥公式即求得
最后是O（C）的查询，c为星星亮度的最大值
代码如下：
#include<bits/stdc++.h> using namespace std; int mp[105][105][11]; int sum[105][105][11]; int main(){ int n,q,c; int x,y,s; int t,x1,y1,x2,y2; while(scanf("%d%d%d",&n,&q,&c) !=EOF){ memset(mp,0,sizeof(mp)); memset(sum,0,sizeof(sum)); for(int i=1;i<=n;i++){ scanf("%d%d%d",&x,&y,&s); mp[x][y][s]++; } for(int k=0;k<=c;k++){ for(int i=1;i<=100;i++){ for(int j=1;j<=100;j++){ sum[i][j][k]=sum[i-1][j][k]+sum[i][j-1][k]-sum[i-1][j-1][k]+mp[i][j][k]; } } } for(int i=0;i<q;i++){ scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2); int ans=0; int ans1; for(int k = 0; k <= c; k++) { ans1 = (k+t)%(c+1); // 计算亮度为k的星星在t秒后的亮度 int num = sum[x2][y2][k]-sum[x1-1][y2][k]-sum[x2][y1-1][k]+sum[x1-1][y1-1][k]; //计算 （x1,y1）到（x2，y2）矩阵内亮度为k的星星的个数 ans1 *= num; ans += ans1; } cout<<ans<<endl; } } }