本题链接：202104-2 邻域均值

本博客给出本题截图：

C++

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 610;

int n, L, r, t;
int s[N][N];

int get_sum(int x1, int y1, int x2, int y2)
{
    return s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
}

int get_cnt(int x1, int y1, int x2, int y2)
{
    return (x2 - x1 + 1) * (y2 - y1 + 1);
}

int main()
{
    scanf("%d%d%d%d", &n, &L, &r, &t);
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
        {
            int x;
            scanf("%d", &x);
            s[i][j] = x + s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
        }

    int res = 0;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
        {
            int x1 = max(1, i - r), y1 = max(1, j - r);
            int x2 = min(n, i + r), y2 = min(n, j + r);
            if (get_sum(x1, y1, x2, y2) <= t * get_cnt(x1, y1, x2, y2))
                res ++ ;
        }
    printf("%d\n", res);
    return 0;
}

  

 

  
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总结

二维前缀和的基础运用.
前缀和

文章来源: chen-ac.blog.csdn.net，作者：辰chen，版权归原作者所有，如需转载，请联系作者。

原文链接：chen-ac.blog.csdn.net/article/details/120103899