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People on Mars count their numbers with base 13:
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
1 #include<stdio.h> 2 #include<string> 3 #include<iostream> 4 #include<string.h> 5 #include<sstream> 6 #include<vector> 7 #include<map> 8 using namespace std; 9 char firwei[13][5]={"tret","jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"}; 10 char secwei[13][5]={"tret","tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"}; 11 map<string,int> mm1,mm2; 12 void toMar(int n) 13 { 14 vector<string> vv; 15 vv.push_back(firwei[n%13]); 16 n = n / 13; 17 if(n > 0) 18 { 19 vv.push_back(secwei[n]); 20 } 21 bool fir = 1; 22 for(int i = vv.size() -1 ; i >= 0 ;--i ) 23 { 24 if(fir) 25 { 26 cout << vv[i]; 27 fir = 0; 28 } 29 else if( vv [i] != "tret" )cout << " " << vv[i] ; 30 } 31 cout << endl; 32 } 33 34 int main() 35 { 36 int n; 37 for(int i = 0 ;i < 13 ;++i) 38 { 39 mm1[firwei[i]] = i; 40 } 41 for(int i = 0 ;i < 13 ;++i) 42 { 43 mm2[secwei[i]] = i; 44 } 45 scanf("%d",&n); 46 getchar(); 47 string str; 48 for(int i = 0 ;i < n;++i) 49 { 50 getline(cin,str); 51 if(str[0]>=‘0‘ && str[0] <= ‘9‘) 52 { 53 int num; 54 stringstream ss; 55 ss << str; 56 ss >> num; 57 toMar(num); 58 } 59 else 60 { 61 stringstream ss; 62 vector<string> svv; 63 string num; 64 int sum = 0; 65 ss << str; 66 while(ss >> num) 67 { 68 svv.push_back(num); 69 } 70 if(svv.size() > 1) 71 { 72 sum = sum*13 + mm2[svv[0]]; 73 sum = sum*13 + mm1[svv[1]]; 74 } 75 else 76 { 77 sum = sum*13 + mm2[svv[0]]; 78 sum = sum*13 + mm1[svv[0]]; 79 } 80 printf("%d\n",sum); 81 } 82 } 83 return 0; 84 }
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原文地址：http://www.cnblogs.com/xiaoyesoso/p/5218871.html
