1 solutions

  • 0
    @ 2025-3-3 16:32:50

    C++ :

    #include <bits/stdc++.h>
    using namespace std;
    
    int main(){
    	int i,j,n,a[20][20] = {0},t=1;
    	cin>>n;
    	for(i = 0;i< n;i++){
    		for(j = 0;j < n - i;j++){
    			a[j][j+i] = t;
    			a[j+i][j] = t;
    			t++;
    		}
    	}
    	
    	for(i = 0;i < n;i++){
    		for(j = 0;j < n;j++){
    			cout<<setw(5)<<a[i][j];
    		}
    		cout<<endl;
    	}
        return 0;
    }
    
    

    Python :

    n = int(input())
    a = [[0 for j in range(n)] for i in range(n)]
    m = 1
    # k认为是左对角线的偏移量
    for k in range(n):
        # i为从上到下的坐标
        for i in range(0, n - k):
            a[i][k + i] = m
            a[k + i][i] = m
            m += 1
    
    # 打印输出
    for i in range(n):
        for j in range(n):
            print('%5d' % a[i][j], end='')
        print()
    
    
    • 1

    Information

    ID
    10400
    Time
    1000ms
    Memory
    16MiB
    Difficulty
    (None)
    Tags
    # Submissions
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