1 条题解

  • 0
    @ 2023-3-13 11:53:10

    C :

    #include<stdio.h>
    int down(int a[200][200], int i, int j,int m) 
    { //方向向下判断
    	if (i < m - 1) 
    	{
    		if (a[i + 1][j] == -1) //如果下一个数据已经去过
    			return 0;
    		else 
    			return 1;
    	} 
    	else if (i == m - 1) //如果到达数组边界
    		return 0;
    	else
    		return 1;
    }
    int right(int a[200][200], int i, int j, int n) 
    {
    	if (j == n - 1)
    		return 0;
    	else if (j < n) 
    	{
    		if (a[i][j + 1] == -1)
    			return 0;
    		else 
    			return 1;
    	}
    }
    int up(int a[200][200], int i, int j) 
    {
    	if (i == 0)
    		return 0;
    	else if (a[i - 1][j] == -1)
    		return 0;
    	else 
    		return 1;
    }
    int left(int a[200][200], int i, int j) 
    {
    	if (j == 0)
    		return 0;
    	else if (a[i][j - 1] == -1)
    		return 0;
    	else 
    		return 1;
    }
    
    int main() 
    {
    	int a[200][200];
    	int i, j;
    	int m, n;
    	scanf("%d%d", &m, &n);
    	for (i = 0; i < m; i++) 
    		for (j = 0; j < n; j++)
    			scanf("%d", &a[i][j]);
    			
    	i = 0;
    	j = 0;
    	
    	while (m!=1 || n!=1) 
    	{
    		for (; down(a, i, j, m); i++) 
    		{
    			printf("%d ", a[i][j]);
    			a[i][j] = -1;//将去过的数据设置标记即-1
    		}
    		
    		if (a[i][j + 1] == -1)
    			break;
    			
    		for (; right(a, i, j, n); j++) 
    		{
    			printf("%d ", a[i][j]);
    			a[i][j] = -1;
    		}
    		
    		if (a[i - 1][j] == -1)
    			break;
    			
    		for (; up(a, i, j); i--) 
    		{
    			printf("%d ", a[i][j]);
    			a[i][j] = -1;
    		}
    		
    		if (a[i][j - 1] == -1)
    			break;
    			
    		for (; left(a, i, j); j--) 
    		{
    			printf("%d ", a[i][j]);
    			a[i][j] = -1;
    		}
    		
    		if (a[i + 1][j] == -1)
    			break;
    	}
    	
    	printf("%d", a[i][j]);
    	return 0;
    }
    
    

    C++ :

    #include <cstdio>  
    #include <vector>  
    #include <cstring>  
    #include <iostream>  
    using namespace std;  
      
    int main()  
    {  
        //freopen("input2(1).txt", "r", stdin);  
        int m, n;  
        int i, j, num = 0;  
        int a[205][205];  
        memset(a, -1, sizeof(a));  
        scanf("%d%d", &m, &n);  
        for(i = 0; i < m; i++)  
            for(j = 0; j < n; j++)  
                scanf("%d", &a[i][j]);  
        int tot = 0, x = -1, y = 0;  
        while(tot < m * n)  
        {  
            while(x + 1 < m && a[x + 1][y] != -1)  
            {  
                printf("%d ", a[++x][y]);  
                a[x][y] = -1;  
                ++tot;  
            }  
            while(y + 1 < n && a[x][y + 1] != -1)  
            {  
                printf("%d ", a[x][++y]);  
                a[x][y] = -1;  
                ++tot;  
            }  
            while(x - 1 >= 0 && a[x - 1][y] != -1)  
            {  
                printf("%d ", a[--x][y]);  
                a[x][y] = -1;  
                ++tot;  
            }  
            while(y - 1 >= 0 && a[x][y - 1] != -1)  
            {  
                printf("%d ", a[x][--y]);  
                a[x][y] = -1;  
                ++tot;  
            }  
        }  
        return 0;  
    }  
    
    • 1

    信息

    ID
    1896
    时间
    1000ms
    内存
    512MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
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