【模拟+读入】#21 A. Jabber ID

A. Jabber ID

time limit per test

2 seconds

memory limit per test

64 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.

There are the samples of correct Jabber IDs: [e-mail],[e-mail]/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

[e-mail]

output

YES

input

[e-mail]/contest.icpc/12

output

NO

这道题是一个给你一个邮箱地址看看合不合规矩的判定：

1.由“A@B/C”的形式组成，A在16个字符内，B在32个字符（不包括.）内，C在16个字符内；

2.可以没有“/C”，但是@的前后都至少要有一个字符

3.B可以由好多个'.'分开，但是每个单词不能超过16个字符。

4.允许的字符在'_' isupper() isdigit() islower() 范围内

我傻乎乎的还想用

scanf("%s@%s/%s",a,b,c);

这样的格式来偷懒……看来是正则表达式看的异想天开了。

AC-Code：

#include <cstdio>  #include <string>#include <cstring>  #include <iostream>  //http://blog.csdn.net/okcd00/article/details/27669883using namespace std;  #define Giveup {cout<<"NO"<<endl;return 0;}char str[101]; bool judge(char c){ return (c=='_'||islower(c)||isdigit(c)||isupper(c)); }int main()  {      //freopen("in.txt","r",stdin);      int cnt=0;      int count=0;//each words - Whole hostname      int type=0;//0-username 1-hostname 2-resource            while(scanf("%s",str)!=EOF)    {type=0;//init is username input        int len=strlen(str);          for(int i=0;i<len;i++)      {          char c=str[i];          //change kind          if(c=='@')          {              if(type==1) Giveup            type=1;              //change to hostname              if(cnt<1||cnt>16) Giveup;            cnt=0;          }          else if(c=='/')          {              if(type==2) Giveup            type=2;//change to resource             if(cnt<1||cnt>16) Giveup            if(count<1||count>32) Giveup            cnt=0;          }          else if(c=='.')          {              if(type!=1) Giveup            if(cnt<1||cnt>16) Giveup            cnt=0;          }          else if(c>127||c<33) Giveup        else if(type==0)         {              if(judge(c))               {                  cnt++;                  continue;              }              else Giveup        }          else if(type==1)          {              if(judge(c))               {                  cnt++;                  count++;                  continue;                 }              else Giveup        }          else if(type==2)          {              if(judge(c))              {                  cnt++;                  continue;              }               else Giveup        }      }      if(cnt<0||cnt>16) Giveup    if(type>=1&&cnt>0) cout<<"YES"<<endl;    else Giveup    }}  

调试用代码:

#include<map>#include<string>#include<cstdio>#include<ctype.h>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define Nx cout<<"NO"char str[101];int main(){//freopen("in.txt","r",stdin);int cnt=0;int count=0;//each words - Whole hostnameint type=0;//0-username 1-hostname 2-resourcetype=0;//init is username inputwhile(scanf("%s",str)!=EOF);int len=strlen(str);for(int i=0;i<len;i++){char c=str[i];//change kindif(c=='@'){if(type==1){Nx;//printf("NO in @");return 0;} type=1;//change to hostnameif(cnt<1||cnt>16){Nx;//printf("NO in @");return 0;} cnt=0;}else if(c=='/'){if(type==2){Nx;//printf("NO in @");return 0;} type=2;//change to resourceif(cnt<1||cnt>16){Nx;//printf("NO in host small");return 0;} if(count<1||count>32){Nx;//printf("NO in host big");return 0;} cnt=0;}else if(c=='.'){if(type!=1) {Nx;//printf("NO in .");return 0;} if(cnt<1||cnt>16){Nx;//printf("NO in .");return 0;} cnt=0;}else if(c>127||c<33)//test if ASCII is right{//if(scanf("%c",&c)!=EOF)  test failed{Nx;//printf("\nNO in range: %c",c);return 0;}}else if(type==0){if(isupper(c)||islower(c)||isdigit(c)||c=='_') {cnt++;continue;}else Nx;//printf("NO in c wrong 0");return 0;}else if(type==1){if(isupper(c)||islower(c)||isdigit(c)||c=='_') {cnt++;count++;continue;}else Nx;//printf("NO in c wrong 1");return 0;}else if(type==2){if(isupper(c)||islower(c)||isdigit(c)||c=='_'){cnt++;continue;} else Nx;//printf("NO in c wrong 2");return 0;}}if(cnt<0||cnt>16){Nx;//printf("NO in resource");return 0;}if(type>=1&&cnt>0) printf("YES");else Nx;return 0;}