A. Boy or Girl

95 阅读 0 评论 63 点赞

我是靠谱客的博主愉快水杯，最近开发中收集的这篇文章主要介绍A. Boy or Girl，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Those days, many boys use beautiful girls' photos as avatars in forums. So it is pretty hard to tell the gender of a user at the first glance. Last year, our hero went to a forum and had a nice chat with a beauty (he thought so). After that they talked very often and eventually they became a couple in the network.

But yesterday, he came to see "her" in the real world and found out "she" is actually a very strong man! Our hero is very sad and he is too tired to love again now. So he came up with a way to recognize users' genders by their user names.

This is his method: if the number of distinct characters in one's user name is odd, then he is a male, otherwise she is a female. You are given the string that denotes the user name, please help our hero to determine the gender of this user by his method.

Input

The first line contains a non-empty string, that contains only lowercase English letters — the user name. This string contains at most 100 letters.

Output

If it is a female by our hero's method, print "CHAT WITH HER!" (without the quotes), otherwise, print "IGNORE HIM!" (without the quotes).

       Sample test(s) 
     
         input 
       
wjmzbmr

         output 
       
CHAT WITH HER!

         input 
       
xiaodao

         output 
       
IGNORE HIM!

         input 
       
sevenkplus

         output 
       
CHAT WITH HER!

Note

For the first example. There are 6 distinct characters in "wjmzbmr". These characters are: "w", "j", "m", "z", "b", "r". So wjmzbmr is a female and you should print "CHAT WITH HER!".

解题说明：把字母映射到字母数组，然后判断是否被2整除即可

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

int main()
{
	int i;
	int sum;
	int po[26];
	char a[102];
	for(i=0;i<26;i++)
	{
		po[i]=0;
	}
	sum=0;
	scanf("%s",&a);
	for(i=0;a[i]!='';i++)
	{
		po[a[i]-'a']=1;
	}
	for(i=0;i<26;i++)
	{
		if(po[i]==1)
		{
			sum++;
		}
	}
	if(sum%2==0)
	{
		printf("CHAT WITH HER!n");
	}
	else
	{
		printf("IGNORE HIM!n");
	}
	return 0;
}