PAT 1100 Mars Numbers

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本文介绍了一种基于地球和火星的计数系统之间的转换程序。火星人使用13进制计数，将地球上的数字0至12分别命名为tret至dec，并为更高位数命名了12个新的名称。通过此程序，可以实现地球和火星数字系统的互译。

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People on Mars count their numbers with base 13:

Zero on Earth is called "tret" on Mars.
The numbers 1 to 12 on Earch is called "jan, feb, mar, apr, may, jun, jly, aug, sep, oct, nov, dec" on Mars, respectively.
For the next higher digit, Mars people name the 12 numbers as "tam, hel, maa, huh, tou, kes, hei, elo, syy, lok, mer, jou", respectively.

For examples, the number 29 on Earth is called "hel mar" on Mars; and "elo nov" on Mars corresponds to 115 on Earth. In order to help communication between people from these two planets, you are supposed to write a program for mutual translation between Earth and Mars number systems.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<100). Then N lines follow, each contains a number in [0, 169), given either in the form of an Earth number, or that of Mars.

Output Specification:

For each number, print in a line the corresponding number in the other language.

Sample Input:

4
29
5
elo nov
tam

Sample Output:

hel mar
may
115
13

#include<cstdio>
#include<set>
#include<map>
#include<string>
#include<iostream>
#include<cctype>
using namespace std;

string gewei[13]={"tret","jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"}; 
string shiwei[13]={"tret","tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"}; 

string NumToStr[170];
map <string,int >StrToNum;
//初始化map 
void init(){
	for(int i=0;i<13;i++){
		NumToStr[i]=gewei[i];
		StrToNum[gewei[i]]=i;
		
		NumToStr[13*i]=shiwei[i];
		StrToNum[shiwei[i]]=13*i;
	}
	for(int i=1;i<13;i++){
		for(int j=1;j<13;j++){
			string str=shiwei[i]+" "+gewei[j];
			NumToStr[i*13+j]=str;
			StrToNum[str]=i*13+j;
		}	
	}		
}

//字符串转化为数字 		
int convert(string s,int &sum){
	sum=0;
	for(int i=0;i<s.length();i++)
		sum=sum*10+s[i]-'0';
	return sum;
} 

int main(){
	init();
	int T;	
	scanf("%d%*c",&T);
	while(T--){
		int sum;
		string s;
		getline(cin,s);
		/*isdigit：数字，isalpha：英文字符，isalnum:数字加英文字符 */ 
		if(isdigit(s[0]))		cout<<NumToStr[convert(s,sum)]<<endl;
		else	cout<<StrToNum[s]<<endl;	
	}
	return 0;
}