PAT (Advanced Level)-1100 Mars Numbers-优快云博客

本文介绍了一个程序，用于在地球数字系统和火星数字系统之间进行转换。火星数字系统采用13进制，将地球上的数字重新映射为独特的火星词汇。文章详细解释了输入输出规格，并提供了一个示例代码实现。

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PAT (Basic Level)-1044 火星数字（20 分）

1100 Mars Numbers （20 分）

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

注意 13的整数倍的情况

#include <iostream>

using namespace std;
int main()
{
	string a[13]={"tret","jan","feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
	string b[13]={"","tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
	int n;
	scanf("%d",&n);
	getchar();
	while(n--)
	{
		string buf;
		getline(cin,buf);
		if(buf[0]>='0'&&buf[0]<='9')
		{
			int num;
			num=stoi(buf);
			if(num>=13)
			{
				if(num%13!=0)cout<<b[num/13]<<" "<<a[num%13]<<endl;
				else cout<<b[num/13]<<endl;
			}
			else cout<<a[num]<<endl;		
		}
		else
		{
			if(buf.size()>3)
			{
				int num=0;
				string high(buf,0,3);
				string low(buf,4,3);
				for(int i=1;i<13;i++)
				{
					if(b[i]==high)num+=i*13;
				}
				for(int i=1;i<13;i++)
				{
					if(a[i]==low)num+=i;
				}
				printf("%d\n",num);
			}
			else
			{
				for(int i=1;i<13;i++)
				{
					if(a[i]==buf)cout<<i<<endl;
					else if(buf==b[i])cout<<i*13<<endl;
				}
				
			}
		}		
	}	
}