HDU 6213 Chinese Zodiac 【模拟/水题/生肖】

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 be 2 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

3 ox rooster rooster ox dragon dragon

 

Sample Output

8 4 12

 

Source

2017 ACM/ICPC Asia Regional Qingdao Online 

 

【题意】:一个女的找了一个比自己小的丈夫，没有人知道他们的年领差，但是知道他们的十二生肖。问女的至少比男的大多少岁。

【分析】如果两人属相相同，肯定大一轮，就是12.

如果女的属相比男的靠后，则就是12-（女的属相-男的属相）

如果女的属相比男的靠前，则就是（男的属相-女的属相）

【代码】：

15MS

1948K

 

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
#include<set>
#include<string>
using namespace std;

int main()
{
    int t;
    string f,m;
    cin>>t;
    map<string,int> mp;
/*rat, ox, tiger, rabbit, dragon, snake, horse, sheep, monkey, rooster, dog and pig.*/
    while(t--)
    {
        cin>>f>>m;
        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;
        if(mp[f]==mp[m]) cout<<12<<endl;
        else if(mp[f]<mp[m]) cout<<abs(mp[f]-mp[m])<<endl;
        else cout<<12-abs(mp[f]-mp[m])<<endl;
    }
    return 0;
}