Star sky 二维前缀和-白红宇

Star sky 二维前缀和

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发布时间：2019-06-14

本文共 3791 字，大约阅读时间需要 12 分钟。

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

Copy

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

Copy

3 0 3

input

Copy

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

Copy

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.

http://codeforces.com/contest/835/problem/C

这题就是暴力二位前缀和

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
              9 #include 
              
               10 #include 
               
                11 #include 
                
                 12 #define pi acos(-1.0)13 #define eps 1e-614 #define fi first15 #define se second16 #define lson l,m,rt<<117 #define rson m+1,r,rt<<1|118 #define bug printf("******\n")19 #define mem(a,b) memset(a,b,sizeof(a))20 #define fuck(x) cout<<"["<
                 
                  <<"]"<
                  
                   = b; i--)29 #define FRL(i,a,b) for(i = a; i < b; i++)30 #define FRLL(i,a,b) for(i = a; i > b; i--)31 #define FIN freopen("DATA.txt","r",stdin)32 #define gcd(a,b) __gcd(a,b)33 #define lowbit(x) x&-x34 #pragma comment (linker,"/STACK:102400000,102400000")35 using namespace std;36 typedef long long LL;37 const int maxn = 1e6 + 10;38 int n, q, c, sum[105][105][12];39 40 int main() {41 sfff(n, q, c);42 for (int i = 0, x, y, s ; i < n ; i++) {43 sfff(x, y, s);44 sum[x][y][s]++;45 }46 for (int i = 1 ; i <= 100 ; i++)47 for (int j = 1 ; j <= 100 ; j++)48 for (int k = 0 ; k <= c ; k++)49 sum[i][j][k] += sum[i - 1][j][k] + sum[i][j - 1][k] - sum[i - 1][j - 1][k];50 int t, x1, y1, x2, y2;51 while(q--) {52 sf(t);53 sffff(x1,y1,x2,y2);54 int ans = 0;55 for (int i = 0 ; i <= c ; i++)56 ans += (sum[x2][y2][i] - sum[x1 - 1][y2][i] - sum[x2][y1 - 1][i] + sum[x1 - 1][y1 - 1][i]) * ((i + t) % (c + 1));57 printf("%d\n", ans);58 }59 return 0;60 }

转载于:https://www.cnblogs.com/qldabiaoge/p/9440141.html

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