杭电 6213 （map） 之 Chinese Zodiac

Problem Description

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 be2 years. But if the signs of the couple is the same, the answer should be 12 years.

 

Input

The first line of input contains an integer 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

3ox roosterrooster oxdragon dragon

 

Sample Output

84
12
题意：一个妹子和她年轻的丈夫年龄都不知道，只知道两个人的生肖，求他们年龄之差的最小值，注意妹子的年纪比她丈夫大就行了
AC代码如下：
#include <iostream>#include <cstring>#include <cstdio>#include <string>#include <map>#include <cmath>#include <algorithm>using namespace std;map<string,int> mp;int main(){    mp["rat"]=1;mp["ox"]=2;mp["tiger"]=3;mp["rabbit"]=4;mp["dragon"]=5;mp["snake"]=6;    mp["horse"]=7;mp["sheep"]=8;mp["monkey"]=9;mp["rooster"]=10;mp["dog"]=11;mp["pig"]=12;    int n;    string str1,str2;    cin>>n;    while(n--)    {        cin>>str1>>str2;        int ans=abs(12+mp[str2]-mp[str1]);        int ans1=abs(12+mp[str2]-mp[str1]);        ans=min(ans,ans1);        if(ans==0) ans=12;        if(ans>12) ans-=12;        cout<<ans<<endl;    }    return 0;}