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));

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<vector>
 #include<algorithm>
 #include<stack>
 #include<cctype>
 using namespace std;
 long long a,b;
 int T;
 long long gcd(long long x,long long y)
 {
     return y== ? x:gcd(y,x%y);
 }
 int main()
 {
     cin>>T;
     while(T--)
     {
         cin>>a>>b;
         long long t1,t2;
         t1=(a+b)/+b;
         t2=*(a+b);
         cout<<t1/(gcd(t1,t2))<<"/"<<t2/(gcd(t1,t2))<<endl;
     }
     return ;
 }