1 solutions

  • 0
    @ 2025-11-5 15:52:18

    C :

    #include<stdio.h>
    #include<string.h>
    int main()
    {
    	int n,i,j,l[10];
    	char x[10][51];
    	scanf("%d",&n);
    	getchar();
    	for(i=0;i<n;i++)
    	{
    		gets(x[i]);
    		l[i]=strlen(x[i]);
    	}
    	for(i=0;i<n;i++)
    	{
    		for(j=0;j<l[i];j++)
    		{
    			if(x[i][j]=='0')
    			x[i][j]='1';
    			else
    			x[i][j]='0';
    		}
    		x[i][j+1]='\0';
    	}
    	for(i=0;i<n;i++)
    	{
    		printf("%s",x[i]);
    		if(i!=n-1)
    		printf("\n");
    	}
    	
    }
    
    • 1

    【设计型】第10章: 字符串 10.27奇偶互换

    Information

    ID
    17274
    Time
    1000ms
    Memory
    2MiB
    Difficulty
    (None)
    Tags
    # Submissions
    0
    Accepted
    0
    Uploaded By