【PAT】A1100 Mars Numbers *

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本文介绍了一种基于13进制的火星计数系统，该系统使用特定的火星月名称来表示数字。通过建立地球数字与火星数字之间的映射，实现了两种计数系统之间的相互转换。文章提供了一个C++实现的例子，包括初始化映射和处理输入输出的过程。

People on Mars count their numbers with base 13:

Zero on Earth is called "tret" on Mars.
The numbers 1 to 12 on Earth 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

解题思路：

理解题意：也是一种进制的转变，但是需要用到映射
代码分为两段：首先是初始化映射，然后根据输入的判断，处理输出。
总结：本题的难点在于初始化的函数，就是整型转变为

代码：

#include <stdio.h>
#include <map>
#include <string>
#include <iostream>
#include <string.h>
using namespace std;
map<string,int> strtonum;
//[0,12]的火星文 
string month[13]={"tret","jan","feb","mar","apr","may","jun","jly","aug","sep","oct","nov","dec"};
//13的[0,12]倍的火星文 
string name[13]={"tret","tam","hel","maa","huh","tou","kes","hei","elo","syy","lok","mer","jou"};

string numtostr[170];

void init(){
	for(int i=0;i<13;i++){
		numtostr[i]=month[i];  //个位为[1,12]，十位为0
		strtonum[month[i]]=i;    //个位数的情况 
		numtostr[i*13]=name[i];    //十位数的情况 
		strtonum[name[i]]=i*13; 
	}
	for(int i=1;i<13;i++){  //十位 
		for(int j=1;j<13;j++){
			string str=name[i]+" "+month[j];  //枚举169个数，并存入字符串数组str容器中
			numtostr[i*13+j]=str;  
			strtonum[str]=i*13+j;
		} 
	}
}

int main(){
	init();
	int T;
	scanf("%d%*c",&T);
	while(T--){
		string str;
		getline(cin,str);
		if(str[0]>='0'&&str[0]<='9'){  //如果是数字 
			int num=0;
			for(int i=0;i<str.length();i++){
				num=num*10+(str[i]-'0');
			}
			cout<<numtostr[num]<<endl;
		}else{
			cout<<strtonum[str]<<endl;
		}
	}
	return 0;
}