A. Jabber ID

2024-08-28 07:18

文章标签 id jabber

本文主要是介绍A. Jabber ID，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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.

Sample test(s) 
input 
mike@codeforces.com
output 
YES
input 
john.smith@codeforces.ru/contest.icpc/12
output 
NO

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long LL;
char s[205];

int dig(char c)
{
    if(c>='0'&&c<='9')return 1;
    return 0;
}
int let(char c)
{
    if(c>='A'&&c<='Z' || c>='a'&&c<='z')return 1;
    return 0;
}

int check()
{
    int n = strlen(s), p = 0;
    while(p<n && s[p]!='@')p++;
    if(p==n)    return 0;
    if(p < 1 || p > 16)return 0;
    //puts("==");

    for(int i = 0; i < p; i++)
        if(s[i]!='_' && !dig(s[i]) && !let(s[i]))return 0;
    //puts("===");
    int pre = p+1;
    while(p<n && s[p]!='/')p++;
    if(p-pre<1 || p-pre>32)return 0;

    int cnt = 0;
    for(int i = pre; i < p; i++)
    {
        if(s[i]=='.')
        {
            if(cnt<1 || cnt>16)return 0;
            cnt = 0;
        }
        else
        {
            if(s[i]!='_' && !dig(s[i]) && !let(s[i]))
                return 0;
            cnt++;
        }
    }
    if(cnt<1 || cnt>16)return 0;

    if(p < n)
    {
    //puts("--");
        cnt = 0;
        for(int i = p+1; i < n; i++)
        {
            if(s[i]=='/')
            {
                if(cnt<1 || cnt>16)return 0;
                cnt = 0;
            }
            else
            {
                if(s[i]!='_' && !dig(s[i]) && !let(s[i]))
                    return 0;
                cnt++;
            }
        }
        if(cnt<1 || cnt>16)return 0;
    }
    return 1;
}
int main()
{
    scanf("%s",s);
    puts(check()?"YES":"NO");

    return 0;
}