1100. Mars Numbers (20)
時間限制
400 ms
內存限制
65536 kB
代碼長度限制
16000 B
判題程序
Standard
作者
CHEN, Yue
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

本題題意就是在題目描述的火星數字規則下與阿拉伯數字的轉換 輸入一個阿拉伯數字 輸出一個對應的火星文 輸入一個火星文 輸出一個對應的地球數字

本題需要注意的是 雖然沒在題目中描述出來 但可以從樣例中看到 當數字是13的整數倍時 不需要輸出0----tret 只需要輸出前面的取整的大數部分

如輸入13時 只需要輸出tam 不需要輸出后面的0 其余情況就是13與10進制的轉換問題了 通過查表和進制轉換輸出結果

本題在數據轉換的時候可以利用stringstream

將string類型轉化為int：

?stringstream s;

?string a;

?cin>>string;

?s<<a;

?int t;

?s>>t;

頭文件：#include<sstream>

后來一直最后兩個數據過不了 。。。查了好久后發現原來是low數組中的數據抄重了一個。。。- -初始化很重要！！！

#include<cstdio> #include<stack> #include<cmath> #include<sstream> #include<cstring> #include<iostream> using namespace std; int main() {int n;//0-24string low[]={ "tret","jan", "feb","mar" , "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov","dec","tam", "hel","maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};cin>>n; getchar();while(n--){string a;getline(cin,a);if(isdigit(a[0])){stringstream s;s<<a;int t=0;s>>t;s.clear();string res;if(t<13)res.append(low[t]);else if(t%13==0)res.append(low[t/13+12]);else{stack<int>s;while(t){s.push(t%13);t/=13; } int d=s.top();s.pop();res.append(low[d+12]);res.append(" ");d=s.top();res.append(low[d]);} cout<<res<<endl;}else{ int res=0;int pos=a.find(" ");if(pos!=-1)for(int i=0;i<=24;i++){int lpos=a.find(low[i]);if(lpos!=-1&&lpos<pos){res+=(i-12)*13;// cout<<(i-12)*13<<endl;}else if(lpos!=-1){res+=i;// cout<<i<<endl;}} else for(int i=0;i<25;i++)if(a.find(low[i])!=-1&&i<13)res+=i;else if(a.find(low[i])!=-1)res+=(i-12)*13;cout<<res<<endl;}}return 0; }

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