PAT(A) 1100

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

模擬題，題目大意就是讓你進制轉換。13進制的轉換，給數字輸出對應的英文，給英文輸出對應的數字。用map 和 string 進行處理就好。

#include
#include
#include
#include
#include 
#include
using namespace std;
int main(){
	int n;
	string A[13]={"0","jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
	string B[13]={"0","tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
	mapa;
	mapb;
	mapa1;
	mapb1;
	for(int i=1;i<=12;i++){
		a.insert(make_pair(A[i],i));
		a1.insert(make_pair(i,A[i]));
	}
	for(int i=1;i<=12;i++){
		b.insert(make_pair(B[i],i*13));
		b1.insert(make_pair(i,B[i]));
	}
	string c;
	cin>>n;
	getchar();
	int index=0;
	for(int i=1;i<=n;i++){
		getline(cin,c);
		if(c[0]>='a'){
			if(c.find(' ',0)!=c.npos||b[c]>=13){
				string c1 = c.substr(0,c.find(' '));//c1 record the higher one
				string c2 = c.substr(c.find(' ')+1,c.size());
				int ans  = b[c1]+a[c2];
				cout<=13){
					int ans = number/13;
					cout<