1100. Mars Numbers (20)

最新推荐文章于 2021-08-27 23:03:28 发布

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本文介绍了一个程序，用于实现地球十进制计数系统与火星十三进制计数系统之间的相互转换。通过建立数字与特定名称间的映射，程序能够准确地将地球上的数字转化为火星形式，反之亦然。

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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

1、我在第一次解题时，采用的是逐条读取、转化的方式。然后读晴神的书，明白还可以先转化好，接下来只用一一对应即可。这种方法无疑效率更好。
2、使用getline之前，注意先用一次getchar()，读取换行符。

#include<iostream>
#include<cstdio>
#include<string>
#include<map>
using namespace std;
string unit[13]={"tret","jan","feb","mar","apr","may","jun","jly","aug","sep","oct","nov","dec"};
string ten[13]={"tret","tam","hel","maa","huh","tou","kes","hei","elo","syy","lok","mer","jou"};
string numToStr[170];
map<string,int> strToNum;
void init(){
    for(int i=0;i<13;i++){
        numToStr[i]=unit[i];
        strToNum[unit[i]]=i;
        numToStr[i*13]=ten[i];
        strToNum[ten[i]]=i*13;
    }
    for(int i=1;i<13;i++){
        for(int j=1;j<13;j++){
            string str=ten[i]+" "+unit[j];
            numToStr[i*13+j]=str;
            strToNum[str]=i*13+j;
        }
    }
}
int main(){
    freopen("in.txt","r",stdin);
    init();
    int T;
    cin>>T;
    getchar();
    while(T--){
        string str;
        getline(cin,str);
        if(str[0]>='0'&&str[0]<='9'){
            int num=0;
            for(int i=0;i<str.size();i++){
                num=num*10+(str[i]-'0');
            }
            cout<<numToStr[num]<<endl;
        }
        else{
            cout<<strToNum[str]<<endl;
        }
    }
    return 0;
}