USACO基础第六题Name That Number

　　这个题简直就是赤裸裸的map，虽然没用过，但是很明显是，也不难，看一眼就会了。因为有可能会给12个数字，所以肯定不能枚举组合吧。文件里面一共才五千个名字，一次循环简直根本花不了什么时间。只要一个字符串一个字符串检查，再决定是否输出就够了。基础题，不多说了。

/*
ID: like_091
PROG: namenum
LANG: C++
*/
#include<iostream>
#include<fstream>
#include<algorithm>
#include<string>
#include<map>
using namespace std;
map<char, char> m;
void fun()
{
	//此处比较脑残
	m['A'] = m['B'] = m['C'] = '2';
	m['D'] = m['E'] = m['F'] = '3';
	m['G'] = m['H'] = m['I'] = '4';
	m['J']  = m['K'] = m['L'] = '5';
	m['M'] = m['N'] = m['O'] = '6';
	m['P'] = m['R'] = m['S'] = '7';
	m['T'] = m['U'] = m['V'] = '8';
	m['W'] = m['X'] = m['Y'] = '9';
}
int main(void)
{
	ifstream fin("namenum.in");
	ofstream cout("namenum.out");
	string s, t;
	fin>>s;
	fun();//预处理
	//用了两个不同的文件
	ifstream cin("dict.txt");
	bool temp = true;
	while (cin>>t)
	{
		//如果长度不同直接下一个
		if (s.length() != t.length())
			continue;
		bool flag = true;
		for (int i = 0; i < t.length(); i++)
			if (m[t[i]] != s[i])
			{
				flag = false;
				break;
			}
		if (flag)
		{
			//猥琐的标记
			temp = false;
			cout<<t<<endl;
		}
	}
	if (temp)cout<<"NONE"<<endl;
	//记住关闭输入流
	fin.close();
	return 0;
}