51nod算法马拉松 contest7

A题

链接：http://www.51nod.com/contest/problem.html#!problemId=1417

推荐链接：http://blog.csdn.net/a837199685/article/details/45009337

设美女取得正面概率是p，反面就是(1-p),就是美女取一正一反和一反一正的概率相同，然后推出公式y=((a+b)/2+b)/(2*(a+b));

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<string>
 5 #include<cmath>
 6 #include<vector>
 7 #include<algorithm>
 8 #include<stack>
 9 #include<cctype>
10 using namespace std;
11 long long a,b;
12 int T;
13 long long gcd(long long x,long long y)
14 {
15     return y==0 ? x:gcd(y,x%y);
16 }
17 int main()
18 {
19     cin>>T;
20     while(T--)
21     {
22         cin>>a>>b;
23         long long t1,t2;
24         t1=(a+b)/2+b;
25         t2=2*(a+b);
26         cout<<t1/(gcd(t1,t2))<<"/"<<t2/(gcd(t1,t2))<<endl;
27     }
28     return 0;
29 }