zoj 计算子矩阵的平方和

做这个题目有点技巧，要算一个矩阵内元素的平方，然后相加，一般的做法超时，此时我们先把一小块一小块的矩阵的平方和算好，保存下来，然后直接求和就好了

H - Evaluate Matrix Sum

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit  Status  Practice  ZOJ 1636

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

计算和的过程需要技巧，就是每一次计算，保证都是成一个矩阵的，比如计算sum[2][1],第二行第一个，得到的不是a[1][1] + a[1][2] + a[1][3] + a[2][1], 而是

a[2][1] + a[1][1];

代码：

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

using namespace std;

int a[ 510 ][ 510 ];
int sum[ 510 ][ 510 ];
int main() {
	int T,n,m,Q;
	int tp;
	int cas=1;
	int r1,c1,r2,c2;
	scanf("%d",&T);
	while( T-- ) {
		scanf("%d%d",&n,&m);
		memset(sum,0,sizeof(sum) );
		for(int i=1;i<=n;i++) {
			tp = 0;
			for(int j=1;j<=m;j++) {
				scanf("%d",&a[i][j]);
				tp += a[i][j] * a[i][j];
				sum[i][j] = sum[ i - 1 ][ j ] + tp;
			}
		}
		scanf("%d",&Q);
		printf("Case %d:\n",cas++);
		while( Q -- ) {
			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;
}