1 solutions

  • 1
    @ 2025-11-7 20:18:33

    本题题解如下: #include <bits/stdc++.h>

    using namespace std;

    double a[1001][1001], b[1001][1001], n, m;

    int main() { cin >> n >> m;

    for (int i = 0; i < n; i++)
    
    	for (int j = 0; j < m; ++j)
    
    		cin >> a[i][j], b[i][j] = a[i][j];
      
    for (int i = 0; i < n; i++)
    
    	for (int j = 0; j < m; j++)
    
    		if (!i || !j || i == n - 1 || j == m - 1)
      
    			b[i][j] = a[i][j];
        
    		else
      
    			b[i][j] = round((a[i][j] + a[i - 1][j] + a[i + 1][j] + a[i][j - 1] + a[i][j + 1]) / 5.0);
        
    for (int i = 0; i < n; ++i) {
    
    	for (int j = 0; j < m; ++j)
    
    		cout << b[i][j] << " ";
      
    	cout << "\n";
    
    }
    
    return 0;
    

    }

    • 1

    Information

    ID
    12069
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    1
    Tags
    # Submissions
    9
    Accepted
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