Code Forces 21 A(模拟)

ShenduCC

发布于 2018-04-26 16:12:12

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发布于 2018-04-26 16:12:12

文章被收录于专栏：算法修养

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.com, 007@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\n");
	else printf("NO\n");

	return 0;
}

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原始发表：2016-04-16 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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