Code Forces 21 A(模拟)

A. Jabber ID

time limit per test

memory limit per test

input

output

<username>@<hostname>[/resource], where

• <username> — is a sequence of Latin letters (lowercase or uppercase), digits or underscores characters «_», the length of<username>
• <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>
• <resource> — is a sequence of Latin letters (lowercase or uppercase), digits or underscores characters «_», the length of<resource>

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

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

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;
}