PAT_A 1100. Mars Numbers (20)

1100. Mars Numbers (20)

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

• 分析
  1、就是要懂题，最开始我就没读明白，英语能力待提高。
  2、这道题很简单，主要是对字符串的处理，比如读取一行getchar() getline()，字符串拆分strtok()

#include<iostream>#include<string>#include<cstring>#include<cstdio>#include<map>using namespace std;map<string,int>m1={     {"tret",0},    {"jan",1},    {"feb",2},    {"mar",3},    {"apr",4},    {"may",5},    {"jun",6},    {"jly",7},    {"aug",8},    {"sep",9},    {"oct",10},    {"nov",11},    {"dec",12},    {"tam",13},//能够整除13的键值对    {"hel",26},    {"maa",39},    {"huh",52},    {"tou",65},    {"kes",78},    {"hei",91},    {"elo",104},    {"syy",117},    {"lok",130},    {"mer",143},    {"jou",156}};map<string,int>m2={    {"tam",1},    {"hel",2},    {"maa",3},    {"huh",4},    {"tou",5},    {"kes",6},    {"hei",7},    {"elo",8},    {"syy",9},    {"lok",10},    {"mer",11},    {"jou",12}};map<int,string>m3={    {0,"tret"},    {1,"jan"},    {2,"feb"},    {3,"mar"},    {4,"apr"},    {5,"may"},    {6,"jun"},    {7,"jly"},    {8,"aug"},    {9,"sep"},    {10,"oct"},    {11,"nov"},    {12,"dec"}};map<int,string>m4={    {1,"tam"},    {2,"hel"},    {3,"maa"},    {4,"huh"},    {5,"tou"},    {6,"kes"},    {7,"hei"},    {8,"elo"},    {9,"syy"},    {10,"lok"},    {11,"mer"},    {12,"jou"}};int main(){    int n=0;    char tmp[110][30];    cin>>n;    //老问题了，c++ primer plus上就专门说过这个问题，没想到在这与到了，要掌握这种读取行的方式    getchar();    string sttmp;    for(int i=0;i<n;i++)    {        getline(cin,sttmp);        memcpy(tmp[i],sttmp.c_str(),sttmp.length()+1);    }    for(int i=0;i<n;i++)    {        if(tmp[i][0]>='0'&&tmp[i][0]<='9')        {            int t=atoi(tmp[i]);            if(t>=13)            {                cout<<m4.at(t/13);                if(t%13!=0)                  cout<<" "<<m3.at(t%13)<<endl;                else                  cout<<endl;            }            else if(t==0)            {              cout<<"tret"<<endl;            }            else             {                  cout<<m3.at(t%13)<<endl;            }        }else        {            int it=0;            //c语言的strtok字符串拆分挺好用的            char *t=strtok(tmp[i]," ");            char *tt=NULL;            if((tt=strtok(NULL," "))!=NULL)            {                it=13*m2.at(t)+m1.at(tt);            }else            {                if(strcmp(t,"tam")==0)                {                  it=13;                }                else//对于能整除13的直接使用高位key-value                  it=m1.at(t);            }            cout<<it<<endl;        }    }    return 0;}