1100. Mars Numbers (20)

1100. Mars Numbers (20)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
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
//由于博客有点小问题，缩进都不见了=。=
#include <iostream>  
#include <math.h>  
#include <string>  
using namespace std;  
int main() {  
string num;  
string num_h;  
int number_10=0;                   //定义十进制的数  
int level,k;  
string lower[13] = { "tret","jan","feb","mar","apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec" };  
string higher[12] = { "tam", "hel", "maa", "huh", "tou", "kes", "hei","elo", "syy", "lok", "mer", "jou" };  
int N,i;  
cin >> N; int temp = 0;  
for (i = 0; i < N; i++) {  
cin >> num;  
if (num[0] >= '0'&&num[0] <= '9') {  
level = num.length();   
number_10 = 0;  
for (k = num.length(); k > 0; k--) {  
number_10 += (num[k - 1] - '0') * pow(10.0, double(num.length() - k));   //string类转化为int  
}  
level = number_10 / 13;                        //转化为13进制  
k = number_10 % 13;  
if (k == 0 && level != 0)                    //3种不同形式的输出  
cout << higher[level - 1] << endl;  
else if (level == 0)  
cout << lower[k] << endl;  
else  
cout << higher[level - 1] <<" " <<lower[k] << endl;  
number_10 = 0;  
}  
else {  
for (k = 0; k < 12; k++) {  
if (num == higher[k]) {  
break;   
}  
}  
if (k!=12) {                              //判断是两位数还是一位数  
if (cin.get() != '\n')              //判断是否换行（最后一位是否为0）  
cin >> num_h;  
else  
num_h = 'tret';  
number_10 = 0;  
for (k = 0; k < 12; k++) {  
if (num == higher[k])  
break;  
}  
k++;  
number_10 += k * 13;  
for (k = 0; k < 13; k++) {  
if (num_h == lower[k])  
break;  
}  
if (k != 13)  
number_10 += k;  
cout << number_10 << endl;  
number_10 = 0;  
}  
else {  
number_10 = 0;  
for (k = 0; k < 13; k++) {  
if (num == lower[k])  
break;  
}
number_10 = k;  
cout << number_10 << endl;  
number_10 = 0;  
}  
}  
}  
}