E - Evaluate Matrix Sum

Description

Given a matrix, the elements of which are all integer number from 0 to 50, you are required to evaluate the square sum of its specified sub-matrix.

Input

The first line of the input contains a single integer T (1 <= T <= 5), the number of test cases.

For each test case, the first line contains two integers m and n (1 <= m, n <= 500), which are the row and column sizes of the matrix, respectively. The next m lines with n numbers each gives the elements of the matrix.

The next line contains a single integer N (1 <= N <= 100,000), the number of queries. The next N lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2 <= m, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner and lower-right corner of the sub-matrix in question.

Output

For each test case, first print the number of the test case, then N lines with one number on each line, the required square sum. Refer to the sample output for details.

Sample Input

2
2 3
1 2 3
4 5 6
2
1 1 2 2
1 3 2 3
3 3
4 2 3
2 5 1
7 9 2
1
1 1 3 3

Sample Output

Case 1:
46
45
Case 2:
193

题意：
给一个n*m矩阵  下表（1,1)~~~(n,m)
给两个点做对角线   求中间矩阵的值
朴素算法也可以过   但是这个在不断输入的同时也将值存在数组中   查询更快

  #include <iostream>
 #include <string.h>
 #include <stdio.h>

 using namespace std;

 int main()
 {
     int t;
     int n,m;
     int k;
     int tt,x1,x2,y1,y2;
     int num;
     int map[505][505];
     scanf("%d",&t);
     for(num=1;num<=t;num++)
     {
         scanf("%d%d",&n,&m);
         memset(map,0,sizeof(map));
         for(int i=1;i<=n;i++)
         {
             for(int j=1;j<=m;j++)
             {
                 scanf("%d",&k);
                 map[i][j]=k*k+map[i-1][j]+map[i][j-1]-map[i-1][j-1];
             }
         }
         printf("Case %d:\n",num);
         scanf("%d",&tt);
         while(tt--)
         {
             scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
             int ans=map[x1-1][y1-1]+map[x2][y2]-map[x1-1][y2]-map[x2][y1-1];
             printf("%d\n",ans);

         }
     }
     return 0;
 }
 
 
 
 
 
 
 
 #include<iostream>
#include <string.h>
#include <stdio.h>

using namespace std;

int a[ 510 ][ 510 ];
int sum[ 510 ][ 510 ];

int main()
{
    int t,n,m,tt;
    int ss;
    int k;
    int r1,c1,r2,c2;
    scanf("%d",&t);
    for(k=1; k<=t; k++)
    {
        scanf("%d%d",&n,&m);
        memset(sum,0,sizeof(sum));
        for(int i=1; i<=n; i++)
        {
            ss = 0;
            for(int j=1; j<=m; j++)
            {
                scanf("%d",&a[i][j]);
                ss += a[i][j] * a[i][j];
                sum[i][j] = sum[ i - 1 ][ j ] + ss;
            }
        }
        printf("Case %d:\n",k);
        scanf("%d",&tt);
        while(tt--)
        {
            scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
            printf("%d\n",sum[ r2 ][ c2 ] - sum[ r1-1 ][ c2 ] -  sum[ r2 ][ c1 - 1 ] + sum[ r1 - 1 ][ c1 - 1 ]  );
        }
    }
    return 0;

}