题目描述
子矩阵的和
s[i,j] = 第i行第j列格子左上部分所有元素的和
(x1,y1)为左上角,(x2,y2)为右下角的子矩阵的和为s[x2,y2] = s[x1-1,y2] - s[x2,y1-1] + s[x1-1,y1-1]
s[i,j]如何计算? s[i,j] = s[i - 1,j] + s[i,j-1] - s[i-1,j-1] + a[i,j]
样例
二维前缀和
算法1
C++ 代码
#include <iostream>
using namespace std;
const int N = 1010;
int n, m, q; // n行m列, q个询问 1 <= q <= 200000
int a[N][N], s[N][N];
int main()
{
scanf("%d %d %d", &n, &m, &q);
for(int i = 1; i <= n; i ++ )
for(int j = 1; j <= m; j ++ )
scanf("%d", &a[i][j]);
for(int i = 1; i <= n; i ++ )
for(int j = 1; j <= m; j ++ )
s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
while(q -- )
{
int x1, y1, x2, y2;
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
printf("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);
}
return 0;
}