题目传送门
Chinese Zodiac
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 495 Accepted Submission(s): 340
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
题目大意：
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，但是知道他们的生肖，问女的至少比男的大多少岁
利用STL的map来做这个题就很方便了。
三种情况：
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+男的属相-女的属相
如果女的属相比男的靠前，则就是 男的属相-女的属相
代码：
#include<iostream>
#include<map>
#include<string>
using namespace std;map<string,int>Map;int main()
{int T;Map["rat"]=1;Map["ox"]=2;Map["tiger"]=3;Map["rabbit"]=4;Map["dragon"]=5;Map["snake"]=6;Map["horse"]=7;Map["sheep"]=8;Map["monkey"]=9;Map["rooster"]=10;Map["dog"]=11;Map["pig"]=12;cin>>T; while(T--){string N,M;cin>>N>>M;int Ca;if(Map[N]>Map[M]){Ca=12+Map[M]-Map[N];}else if(Map[N]==Map[M]){Ca=12;}else{Ca=Map[M]-Map[N];}printf("%d\n",Ca);}
}