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 tamSample Output:
hel mar may 115 13map的魅力应该在于建立字符与整数之间的映射关系 或者是字符串与字符串之间的映射
#include <bits/stdc++.h> using namespace std; const int N = 1010; //map好用的地方应该是把字符串整成数字 //0-12的火星文 string low[13] = {"tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"}; //13的0-12的 string high[13] = {"tret", "tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"}; string numtostr[170]; //数字->火星文 map<string, int> strtonum; //火星文->数字 void init() { for(int i = 0; i < 13; i++) { numtostr[i] = low[i]; //个位为[0,12],十位为0 strtonum[low[i]] = i; numtostr[i * 13] = high[i]; strtonum[high[i]] = i * 13; } for(int i = 1; i < 13; i++) //十位 { for(int j = 1; j < 13; j++) //个位 { string str = high[i] + " " + low[j]; //火星文 numtostr[i * 13 + j] = str; //数字->火星文 strtonum[str] = i * 13 + j; //火星文->数字 } } } int main() { init(); int n; cin >> n; getchar(); for(int i = 0; i < n; i++) { string s; getline(cin, s); if(s[0] >= '0' && s[0] <= '9') { int num = 0; for(int i = 0; i < s.length(); i++) { num = num * 10 + (s[i] - '0'); } cout << numtostr[num] << endl; //直接查表 } else { cout << strtonum[s] << endl; //直接查表 } } return 0; }