IP判断 （C语言代码）

解题思路:
利用ASCII解题
注意事项:

参考代码:

#include"stdio.h"
#include"string.h"
int main()
{
    
    char ch[30];//存放不超过30个字符// 
    int i,j,l;  //定义用于循环的变量//
    int o,p;   //定义用于判断的变量//
    char n[4][4];
    int temp;
    while(scanf("%s",ch)!=EOF) 
    {    
        temp=0;
        p=0;
        j=0;
        l=0;
        o=0; //初始化// 
        
        for(i=0;i<strlen(ch);i++)  //遍历字符// 
        {
            if(ch[i]<='9'&&ch[i]>='0')
                o=0;
            else if(ch[i]=='.')
                o=0;
            else
            {
                o=1; //为了不满足下面的if条件// 
                printf("N\n"); 
                break;
             }
        }
        
        if(o==0)
        {    
            for(i=0;i<strlen(ch);i++)  //遍历字符//
            {
                if(ch[i]!='.')
                {
                    n[j][l]=ch[i];
                    l++;
                }
                else
                {    n[j][l]='\0'; //添加结束符，为了使用strlen// 
                    l=0; //从下一行的开头开始// 
                    j++;
                }
            }//拆，取数字// 
            
            for(j=0;j<4;j++)
            {    
                if(strlen(n[j])==3)   //只用判断三位数是否满足条件// 
                {
                    temp=(n[j][0]-'0')*100;
                    temp=temp+(n[j][1]-'0')*10;
                    temp=temp+(n[j][2]-'0');
                }  
                if(temp>255)   //只要有一个不满足条件就直接 N // 
                {    
                    p=1;
                    printf("N\n");
                    break;
                }
            }
            if(p!=1)
                printf("Y\n");  //历经千辛万苦，终于满足了 Y的条件，恭喜// 
        }
    }
    return 0;
}