PAT 1100 Mars Numbers[难]

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 N (<100). Then Nlines 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

题目大意：第二点给出的是1-12对应的，第三点给出的是更高位。

//感觉好奇怪，13为什么后面没有0呢？直接输出那样，不太理解。

代码转自：https://www.liuchuo.net/archives/1892

#include <iostream>
#include <string>
#include <cctype>
#include<string.h>
#include<cstdio>
using namespace std;
string a[] = { "tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec" };
string b[] = { "", "tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou" };
void func1(string s) {
    int len = s.length(), num = ;
    for (int i = ; i < len; i++)
        num = num *  + (s[i] - '');
    if (num / ) {
        cout << b[num / ];
        if (num % ) cout << ' ' << a[num % ];
    } else {
        cout << a[num % ];
    }
}
void func2(string s) {
    int len = s.length(), num = ;
    if (len == ) {
        cout << ;
        return;
    } else if (len == ) {//如果是一位。
        for (int i = ; i <= ; i++) {
            if (s == a[i]) {
                cout << i;
                return;
            }
            if (s == b[i]) {
                cout << i * ;
                return;
            }
        }
    }
    else {
        string temp1 = s.substr(, ), temp2 = s.substr(, );
        for (int i = ; i <= ; i++) {//将其转换为十进制。
            if (temp1 == b[i]) num += i * ;
            if (temp2 == a[i]) num += i;
        }
        cout << num;
    }
    return;
}
int main() {
    int n;
    cin >> n;
    getchar();
    for (int i = ; i < n; i++) {
        string s;
        getline(cin, s);
        if (isdigit(s[]))
            func1(s);
        else
            func2(s);
        cout << endl;
    }
    return ;
}