1100. Mars Numbers (20)

最新推荐文章于 2020-10-30 21:21:25 发布

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最新推荐文章于 2020-10-30 21:21:25 发布

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本文介绍了一个程序，用于实现地球十进制计数系统与火星十三进制计数系统之间的相互转换。火星人使用独特的计数方式，将0至12的数字分别命名为tret至dec，对于更大的数字，则采用tam至jou进行命名。该程序能够处理输入文件中的多个测试案例，并准确地将地球数字转换为火星数字，反之亦然。

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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<iostream>  
#include<algorithm>
#include<cstring>  
using namespace std; 
string a1[13]= {"tret","jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
string a2[13]= {"tam","hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
int n,s1,g,j,sum,i,h;
int main()  
{  
cin>>n;
string s,ss="";
getchar();
for(i=1;i<=n;i++)
{
  getline(cin,s);
  if(s[0]>='0'&&s[0]<='9')
  {
       sum=0;
       for(j=0;j<s.size();j++)
           sum=sum*10+s[j]-'0';
    if(sum<=12)
    cout<<a1[sum]<<endl;
   else
     {
        if(sum%13==0)  cout<<a2[sum/13-1]<<endl;
        else cout<<a2[sum/13-1]<<" "<<a1[sum%13]<<endl;
      }
  }
 else
 {  sum=0; 
     for(int l=0;l<s.size();l=l+4)
     {
        ss=ss+s[l]+s[l+1]+s[l+2];
        for(int r=0;r<13;r++)
        {
            if(ss==a1[r])
               sum=sum+r;
             if(ss==a2[r])
               sum=sum+(r+1)*13;
         }
           ss="";
     } 
         cout<<sum<<endl;
  }  
  }  
    return 0;  
}