2017-5-14 湘潭市赛 Strange Optimization

Strange Optimization
Accepted :            Submit :
Time Limit :  MS           Memory Limit :  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(+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.

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

Output

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

Sample Output

/
/

Note

For the first sample, α= 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（i,j都为整数），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。令Xk=k*d/(n*m)

那么相邻两个结果之间的距离为Xk+1-Xk=d/(n*m)， 问题转化为a这个位置到最近的Xk(k为整数)的距离要最大，a应该为Xk+1,Xk（k为整数）的中间位置，这样a到最近的Xk（k为整数）距离最大为d/(2*n*m)。

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+;
LL gcd(LL a,LL b)
{
    return b==?a:gcd(b,a%b);
}
int main()
{
    LL n, m;
    while(scanf("%I64d%I64d",&n,&m)==)
    {
        printf("1/%I64d\n",n/gcd(n,m)*m*);
    }
    return ;
}