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  • 1100 Mars Numbers 字符串操作

    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 (<). 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
    
    #include<iostream>
    #include<map>
    #include<string>
    using namespace std;
    string s;
    int m;
    string a[13] = {"tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
    string b[13] = {"####", "tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
    void f1()
    {
        int res=0,i=0;
        int n=s.size();
        while(i<n)
        {
            res=res*10+(s[i++]-'0');
        }
        int t=res;
        if (t / 13) cout << b[t / 13];
        if ((t / 13) && (t % 13)) cout << " ";
        if (t % 13 || t == 0) cout << a[t % 13];
    }
    void f2()
    {
        string s1,s2;
        int t1=0,t2=0;
        s1=s.substr(0,3);
        int n=s.size();
        if(n>4)
        s2=s.substr(4,3);
        for (int j = 1; j <= 12; j++) {
            if (s1 == a[j] || s2 == a[j]) t2 = j;
            if (s1 == b[j]) t1 = j;
        }
        cout << t1 * 13 + t2;
    }
    int main()
    {
        cin>>m;
        getchar();
        while(m--)
        {
        getline(cin,s);
        if('0'<=s[0]&&s[0]<='9')
        f1();
        else
        f2();
        cout<<endl;
       }
        return 0;
     } 
    如果你够坚强够勇敢,你就能驾驭他们
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  • 原文地址:https://www.cnblogs.com/liuzhaojun/p/11221287.html
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