思路：看了很久才知道题意，虽然51nod分类为博弈，但我感觉还是不像。解法已经隐藏在样例中了，无论正还是反，最后的期望都是一样的。设美女正面的概率为p,反面为（1-p）则
p*A-（1-p）*（A+B）/2 = （1-p）*B-p*（A+B）/2，解得：P=（A+3*B）/（4A+4B），最后取下gcd求最简即可。
1 #include <iostream> 2 #include <queue> 3 #include <stack> 4 #include <cstdio> 5 #include <vector> 6 #include <map> 7 #include <set> 8 #include <bitset> 9 #include <algorithm> 10 #include <cmath> 11 #include <cstring> 12 #include <cstdlib> 13 #include <string> 14 #include <sstream> 15 #include <time.h> 16 #define x first 17 #define y second 18 #define pb push_back 19 #define mp make_pair 20 #define lson l,m,rt*2 21 #define rson m+1,r,rt*2+1 22 #define mt(A,B) memset(A,B,sizeof(A)) 23 #define mod 1000000007 24 using namespace std; 25 typedef long long LL; 26 const double PI = acos(-1); 27 const int N=2e5+10; 28 const int inf = 0x3f3f3f3f; 29 const LL INF=0x3f3f3f3f3f3f3f3fLL; 30 LL gcd(LL a,LL b) 31 { 32 return b==0?a:gcd(b,a%b); 33 } 34 int main() 35 { 36 #ifdef Local 37 freopen("data.txt","r",stdin); 38 #endif 39 int T; 40 LL a,b; 41 cin>>T; 42 while(T--) 43 { 44 cin>>a>>b; 45 LL p=a+3*b; 46 LL q=4*(a+b); 47 LL k=gcd(p,q); 48 cout<<p/k<<"/"<<q/k<<endl; 49 } 50 #ifdef Local 51 cerr << "time: " << (LL) clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl; 52 #endif 53 }