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  • Code Forces 21 A(模拟)

    A. Jabber ID
    time limit per test
    0.5 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Jabber ID on the national Berland service «Babber» has a form <username>@<hostname>[/resource], where

    • <username> — is a sequence of Latin letters (lowercase or uppercase), digits or underscores characters «_», the length of<username> is between 1 and 16, inclusive.
    • <hostname> — is a sequence of word separated by periods (characters «.»), where each word should contain only characters allowed for <username>, the length of each word is between 1 and 16, inclusive. The length of <hostname> is between 1 and 32, inclusive.
    • <resource> — is a sequence of Latin letters (lowercase or uppercase), digits or underscores characters «_», the length of<resource> is between 1 and 16, inclusive.

    The content of square brackets is optional — it can be present or can be absent.

    There are the samples of correct Jabber IDs: mike@codeforces.com007@en.codeforces.com/contest.

    Your task is to write program which checks if given string is a correct Jabber ID.

    Input

    The input contains of a single line. The line has the length between 1 and 100 characters, inclusive. Each characters has ASCII-code between 33 and 127, inclusive.

    Output

    Print YES or NO.

    Examples
    input
    mike@codeforces.com
    
    output
    YES
    
    input
    john.smith@codeforces.ru/contest.icpc/12
    
    output
    NO
    
    直接模拟
    #include<stdio.h>
    #include<string.h>
    bool panduan(char c)
    {
    	if((c!='_')&&(c<'a'||c>'z')&&(c<'A'||c>'Z')&&(c<'0'||c>'9'))return false;
    	else return true;
    }
    int main()
    {
    	char s[100+10];
    	gets(s);
    
    
    
    	int len=strlen(s);
    	char user[100];
    	char host[100];
    	char res[100];
    	int i,len1,len2,len3;
    	int ans=1;
    	//user
    	for(i=0;i<len;i++)
    	{
    		if(s[i]=='@')
    		{
    			len1=i;
    			break;
    		}
    		user[i]=s[i];
    		if(i==len-1&&s[i]!='@')ans=0;
    	}
    	if(s[len-1]=='@')
            ans=0;
    	if(ans==1)
    	{
    		if(len1<1||len1>16)ans=0;
    		else
    		{
    			for(i=0;i<len1;i++)
    			{
    				if(!panduan(user[i]))
    				{
    					ans=0;
    					break;
    				}
    			}
    		}
    	}
    	//host
    	for(i=len1+1;i<len;i++)
    	{
    		if(s[i]=='/')
    		{
    			len2=i-len1-1;
    			break;
    		}
    		host[i-len1-1]=s[i];
    		if(i==len-1&&s[i]!='/')
    		{
    			len2=len-len1-1;
    		}
    	}
    	if(ans==1)
    	{
    		if(len1<1||len1>32)ans=0;
    		else
    		{
    			int sum=0;
    			for(i=0;i<len2;i++)
    			{
    				if(!panduan(host[i]))
    				{
    					if(host[i]=='.')
                        {
                            if(sum>16||sum<1||(i==len2-1))
    				        {
    					       ans=0;break;
    				        }
                            sum=0;
                        }
    					else ans=0;
    				}
    				else sum++;
    				if(ans==0)
                        break;
    
    			}
    		}
    	}
    	//res
    	if(s[len-1]=='/')
            ans=0;
    	if(len1+len2+2<len&&ans==1)
    	{
    		for(i=len1+len2+2;i<len;i++)
    		res[i-len1-len2-2]=s[i];
    		len3=len-2-len1-len2;
    		if(len3<1||len3>16)ans=0;
    		else{
    		for(i=0;i<len3;i++)
    		{
    			if(!panduan(res[i]))
    			{
    				ans=0;
    				break;
    			}
    		}}
    	}
    
    	if(ans==1)printf("YES
    ");
    	else printf("NO
    ");
    
    	return 0;
    }
    



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  • 原文地址:https://www.cnblogs.com/dacc123/p/8228721.html
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