PTA 1100 Mars Numbers

最新推荐文章于 2021-08-12 19:18:26 发布

今天也要加油鸭！！！！！

最新推荐文章于 2021-08-12 19:18:26 发布

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分类专栏： OJ

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本文介绍了一个程序，用于在地球的十进制数字系统和火星的十三进制数字系统之间进行转换。火星数字系统采用独特的命名方式，包括'tret'表示零，'jan'到'dec'表示1到12，以及'tam'到'jou'表示更高的数字。

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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

题目大意：

数字转换，把十进制数字转化为火星文，或者把火星文转化为十进制数字【具体火星文上面有给出，三行分别是0，低位火星文1-12（可以理解为个位数），高位火星文1-12（就类似于235中的2和3）】

题目解析：

这个代码写起来有点啰嗦，被我改来改去我的代码已经改成有点诡异的样子了，有一个坑，就是注意一下当输入tam的时候应该输出13，而当输入13的时候也应该输出tam而不是tam zret

贴代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include<algorithm>
#include <cmath>
#include<queue>
#include<map>
#include<stack>
#include<sstream>
using namespace std;

int main()
{
    char low[13][13] = {"tret","jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
    char high[13][13] = {"tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
    int n;
    string s;
    cin >> n;
    stringstream ss;
    int t;
    getchar();
    for(int i= 0; i < n; i ++)
    {
        getline(cin,s);
        int ans = 0;
        int l = s.length();
        ss << s;
        if('0' <= s[0] && s[0] <= '9')
        {
            ss >> t;
            //cout << t << endl;

            if(t < 13)cout << low[t] <<endl;
            else
            {
                cout << high[t/13-1] ;
                t%=13;
                if(t!=0)
                cout << " " << low[t] ;
                cout << endl;
            }
            

        }
        else
        {
            ss >> s;
            int j;
            if(l <= 4)
            {
                for(j = 0; j < 13; j ++)
                {
                    int k = 0;
                    for(k = 0; k < s.length(); k ++)
                    {
                        if(s[k] != low[j][k])break;
                    }
                    if(k == s.length())
                    {
                        cout << j << endl;
                        break;
                    }
                }
                for(j = 0; j < 13; j ++)
                {
                    int k = 0;
                    for(k = 0; k < s.length(); k ++)
                    {
                        if(s[k] != high[j][k])break;
                    }
                    if(k == s.length())
                    {
                        cout << (j+1)*13 << endl;
                        break;
                    }
                }
            }
            else
            {
                for(j = 0; j < 13; j ++)
                {
                    int k = 0;
                    for(k = 0; k < s.length(); k ++)
                    {
                        if(s[k] != high[j][k])break;
                    }
                    if(k == s.length())
                    {
                        ans = (j+1)*13;
                        break;
                    }
                }
                ss >> s;
                for(j = 0; j < 13; j ++)
                {
                    int k = 0;
                    for(k = 0; k < s.length(); k ++)
                    {
                        if(s[k] != low[j][k])break;
                    }
                    if(k == s.length())
                    {
                        cout << ans + j << endl;
                        break;
                    }
                }
            }
        }
        ss.clear();
    }
    return 0;
}