1100. Mars Numbers (20)

時間限制 400 ms 內存限制 65536 kB 代碼長度限制 16000 B 判題程序 Standard 作者 CHEN, Yue

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

解析：略 #include <iostream> #include <string> #include <cstring> #include <vector> #include <map> #include <stdio.h> using namespace std;string dict[13] = {"tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"}; string dict2[13] = {"", "tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};map<string, int> table; map<string, int> table2;void init(){for(int i=0; i<13; i++){table[dict[i]] = i;}for(int i=0; i<13; i++){table2[dict2[i]] = i;}return; }int pow(int n, int e){if(e == 0){return 1;}else{return n * pow(n, e-1);} }int mar2ear(string input){vector<string> tmp;int len = input.length();string str;for(int i=0; i<len; i++){if(input[i] != ' '){str.push_back(input[i]);if(i == len-1){tmp.push_back(str);str.clear();}}else{tmp.push_back(str);str.clear();}}int result = 0;for(int i=0; i<tmp.size(); i++){if(table[tmp[i]]){result += table[tmp[i]];}else{result += table2[tmp[i]] * 13;}}return result; }string ear2mar(int n){string result;int tmp1 = n/13;int tmp2 = n%13;if(tmp1 == 0){result = dict[tmp2];}else{result.append(dict2[tmp1]);if(tmp2 != 0){result.append(" ");result.append(dict[tmp2]);}}return result; }int toint(string input){int result = 0;int pos = 0;for(int i=input.length()-1; i>=0; i--){result += (input[i] - '0') * pow(10, pos);pos++;}return result; }int main(){init();int n;scanf("%d", &n);getchar();string line;while(n--){getline(cin, line);//is digitif(line[0] >= '0' && line[0] <= '9'){cout<<ear2mar(toint(line))<<endl;}else{cout<<mar2ear(line)<<endl;}}return 0; }

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轉載于:https://www.cnblogs.com/RookieCoder/p/5075288.html

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