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pat 1100 Mars Numbers（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

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

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <string>
  6 #include <map>
  7 #include <stack>
  8 #include <vector>
  9 #include <queue>
 10 #include <set>
 11 #define LL long long
 12 #define INF 0x3f3f3f3f
 13 using namespace std;
 14 const int MAX = 30;
 15 
 16 int n;
 17 string s;
 18 char s1[MAX][6] = {"tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
 19 char s2[MAX][6] = {" ", "tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
 20 
 21 int my_pow(int x, int n)
 22 {
 23     int ans = 1;
 24     while (n)
 25     {
 26         if (n & 1) ans *= x;
 27         x *= x;
 28         n >>= 1;
 29     }
 30     return ans;
 31 }
 32 
 33 int main()
 34 {
 35 //    freopen("Date1.txt", "r", stdin);
 36     scanf("%d", &n);
 37     getchar();
 38     for (int i = 1; i <= n; ++ i)
 39     {
 40         getline(cin, s);
 41         if (isalpha(s[0]))
 42         {
 43             int len = s.size();
 44             if (len >= 6)
 45             {
 46                 char str1[6], str2[6];
 47                 int a = 0, b = 0, ans = 0;
 48                 for (a, b; a < len; ++ a, ++ b)
 49                 {
 50                     if (s[a] == ' ') break;
 51                     str1[b] = s[a];
 52                 }
 53                 str1[b] = '';
 54                 ++ a, b = 0;
 55                 for (a, b; a < len; ++ a, ++ b)
 56                     str2[b] = s[a];
 57                 str2[b] = '';
 58                 for (int i = 0; i < 13; ++ i)
 59                 {
 60                     if (strcmp(s1[i], str2) == 0)
 61                     {
 62                         ans += i;
 63                         break;
 64                     }
 65                 }
 66                 for (int i = 1; i <= 13; ++ i)
 67                 {
 68                     if (strcmp(s2[i], str1) == 0)
 69                     {
 70                         ans += i * 13;
 71                         break;
 72                     }
 73                 }
 74                 printf("%d
", ans);
 75             }
 76             else
 77             {
 78                 char str1[10];
 79                 for (int i = 0; i < len; ++ i)
 80                     str1[i] = s[i];
 81                 str1[len] = '';
 82                 bool flag = false;
 83                 for (int i = 0; i < 13; ++ i)
 84                 {
 85                     if (strcmp(str1, s1[i]) == 0)
 86                     {
 87                         flag = true;
 88                         printf("%d
", i);
 89                         break;
 90                     }
 91                 }
 92                 if (flag) continue;
 93                 for (int i = 1; i < 13; ++ i)
 94                 {
 95                     if (strcmp(str1, s2[i]) == 0)
 96                     {
 97                         printf("%d
", i * 13);
 98                         break;
 99                     }
100                 }
101             }
102         }
103         else
104         {
105             int ans = 0, len = s.size(), t1, t2;
106             for (int i = 0, j = len - 1; i < len; ++ i, -- j)
107                 ans += (s[i] - '0') * my_pow(10, j);
108             t1 = ans / 13, t2 = ans % 13;
109             if (ans <= 12)
110                 printf("%s
", s1[ans]);
111             else if (t2 == 0)
112                 printf("%s
", s2[t1]);
113             else
114                 printf("%s %s
", s2[t1], s1[t2]);
115         }
116     }
117     return 0;
118 }

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原文地址：https://www.cnblogs.com/GetcharZp/p/9595351.html