【Codeforces Round #427】 C 【打表+DP】

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.

Example
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.
好不容易思路对了，但是bug好多，调了老半天才过，zz 啊。
一看就只能够打表来。
我们用dp [ k ] [ i ][ j ] 来表示 时间为k的时候，从[1,1]到[ i , j ]得到的总亮度。
然后我们就可以用这个来表示任意的区域总和为多少。
画个图，就可以找到规律了

#include<bits/stdc++.h>using namespace std;#define  LL long long#define fread() freopen("in.txt","r",stdin)#define fwrite() freopen("out.txt","w",stdout)#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 1e5+10;const int MAXM = 1e6;const int mod = 1e9+7;const int inf = 0x3f3f3f3f;int mp[15][110][110];int main(){    CLOSE();//  fread();//  fwrite();    int n,q,c;    scanf("%d%d%d",&n,&q,&c);    while(n--){        int x,y,val;        scanf("%d%d%d",&x,&y,&val);        for(int i=0; i<=10; i++)            mp[i][x][y] += (val+i)%(c+1);    }    for(int k=0; k<=10; k++){   // 因为取完模之后，其实最多就只有11中情况         for(int i=1; i<=100; i++)            for(int j=1; j<=100; j++)                mp[k][i][j] += mp[k][i-1][j]+mp[k][i][j-1]-mp[k][i-1][j-1];     } //  puts("");//  for(int i=1;i<=3;i++){//      for(int j=1;j<=3;j++)//          printf("%d ",mp[2][i][j]);//      puts("");//  }    while(q--){        int t,x1,y1,x2,y2;        scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);          t %= (c+1);         x1--;y1--; // 因为是左右区间都是闭的，所以前面小的减1就好。         printf("%d\n",mp[t][x2][y2]-mp[t][x1][y2]-mp[t][x2][y1]+mp[t][x1][y1]);     }    return 0;}