Chinese Zodiac （水题）

The Chinese Zodiac, known as Sheng Xiao, is based on a twelve-year cycle, each year in the cycle related to an animal sign. These signs are the rat, ox, tiger, rabbit, dragon, snake, horse, sheep, monkey, rooster, dog and pig. 

Victoria is married to a younger man, but no one knows the real age difference between the couple. The good news is that she told us their Chinese Zodiac signs. Their years of birth in luner calendar is not the same. Here we can guess a very rough estimate of the minimum age difference between them. 

If, for instance, the signs of Victoria and her husband are ox and rabbit respectively, the estimate should be 22 years. But if the signs of the couple is the same, the answer should be 1212 years.

Input

The first line of input contains an integer T (1≤T≤1000)T (1≤T≤1000) indicating the number of test cases. 

For each test case a line of two strings describes the signs of Victoria and her husband.

Output

For each test case output an integer in a line.

Sample Input

3
ox rooster
rooster ox
dragon dragon

Sample Output

8
4
12

题解：感觉用map更好，但是我STL玩的还是不是特别好，就用二维字符数组代替了

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<map>

using namespace std;

int main()
{
   char a[12][15]={"rat","ox","tiger","rabbit","dragon","snake","horse","sheep","monkey","rooster","dog","pig"};
   int n;
   cin>>n;
   char s1[15],s2[15];
   int k1,k2;
   for(int t=0;t<n;t++)
   {

   	  scanf("%s%s",s1,s2);
   	  for(int j=0;j<12;j++)
   	  {
   	  	if(strcmp(s1,a[j])==0)
   	  	{
   	  	    k1=j;
		}
		if(strcmp(s2,a[j])==0)
		{
			k2=j;
		}
	  }
	  if(k1==k2)
	  {
	  	printf("12\n");
	  }
	  if(k1<k2)
	  {
	  	printf("%d\n",k2-k1);
	  }
	  if(k1>k2)
	  {
	  	printf("%d\n",12+k2-k1);
	  }
   }
	return 0;
}