题目描述

样例

源代码

#include<iostream>
#include<cstring>
using namespace std;
const int N = 500;
int flag[N + 1][N + 1];
int n;
int main()
{
    
	
	cin >> n;
	int a=1, b=1;
	int i, j;
	for (i = 1; i < n + 1; i++)
	{
    
		for (j = 1; j < n + 1; j++)
		{
    
			cin>>flag[i][j];
		}
	}
			printf("%d ", flag[a][b]);
			while (a <= n && b <= n)
			{
    
				if (b < n)
				{
    
					b++;
					if (flag[a][b])
					{
    
						printf("%d ", flag[a][b]);
					}
					
				}

				else if (b == n)
				{
    
					a++;
					if (flag[a][b])
					{
    
						printf("%d ", flag[a][b]);
					}
				}
				while (b > 1&& a<n)
				{
    
					b--;
					a++;
					if (flag[a][b])
					{
    
						printf("%d ", flag[a][b]);
					}
				}
				if (a < n)
				{
    
					a++;
					if (flag[a][b])
					{
    
						printf("%d ", flag[a][b]);
					}
				}
				else if (a == n)
				{
    
					b++;
					if (flag[a][b])
					{
    
						printf("%d ", flag[a][b]);
					}
				}
				while (a > 1 && b<n)
				{
    
					a--;
					b++;
					if (flag[a][b])
					{
    
						printf("%d ", flag[a][b]);
					}
				}
				
			}														
	return 0;
}

关于这题

这里 I took a stupid approach
topic actually except for the first number The rest is a loop pattern
①If you can go right 往右 到头了 则往下
②斜向下
③can go down 则往下 到头了 则往右
④斜向上
我们把数组初始化为0 如果超过了 The given range is not printed