Strange Optimization
Accepted : 35           Submit : 197
Time Limit : 1000 MS           Memory Limit : 65536 KB

Strange Optimization

Bobo is facing a strange optimization problem. Given n,m, he is going to find a real number α such that f(12+α) is maximized, where f(t)=mini,j∈Z|in?jm+t|. Help him!

Note: It can be proved that the result is always rational.
Input

The input contains zero or more test cases and is terminated by end-of-file.

Each test case contains two integers n,m.

    1≤n,m≤109
    The number of tests cases does not exceed 104.

Output

For each case, output a fraction p/q which denotes the result.
Sample Input

1 1
1 2

Sample Output

1/2
1/4

Note

For the first sample, α=0 maximizes the function.

Source
XTU OnlineJudge 

/**
题目：Strange Optimization
链接：http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1268
题意：如题目所述。
思路：
f(1/2+a) ,a可以为任意实数，所以实际上等价于f(a)； a为任意实数;

i/n - j/m = (m*i-n*j)/(n*m); 分子看上去像是一个ax+by=c这样的式子，也就是x，y有解，那么c一定是gcd(a,b)的倍数。

所以m*i-n*j = k*gcd(n,m)； k为整数。原式转化为 min |k*d/(n*m) + a| 中的最大值。

那么相邻两个结果之间的距离为d/(n*m)， 所以要想值最小且最大，应该让t的值为中点。那么d/(2*n*m)。那么点距离中点最小的距离为d/(2*n*m).是最大的最小解。

ans = 1/(2*n*m/gcd(n,m)) ;

*/

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> P;
const int maxn = 1e5+100;
LL gcd(LL a,LL b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    LL n, m;
    while(scanf("%I64d%I64d",&n,&m)==2)
    {
        printf("1/%I64d\n",n/gcd(n,m)*m*2);
    }
    return 0;
}