Розбір

Let’s consider the values that the expression $f(i,j)=i/n-j/m$ takes, where $i,j\in Z$ for fixed $n$ and $m$. For example, $f(n,m)=n/n-m/m=0$.

Let’s consider, for instance, the function $f(i,j)=i/3-j/5$. Then

According to the extended Euclidean algorithm, there always exist integers $i$ and $j$, such that $i/3-j/5=1$. Therefore, the smallest positive value that the expression $i/3-j/5$ can take is $1/15$. From this, it follows that this expression can take all possible values of the form $k/15$, where $k\in Z$.

Theorem. If $f(i,j)=i/n-j/m$, then this function takes values of the form $k/LCM(n,m)$, where $k\in Z$.

Proof. Indeed,

Since the numbers $m/GCD(n,m)$ and $n/GCD(n,m)$ are coprime, there always exists integers $i$ and $j$ such that the numerator of the fraction equals $1$. Therefore, the function $f(i,j)$ can take all possible values of the form $k/LCM(n,m)$, where $k\in Z$.

Now let's consider the original function

Its values are the distances from the point $t$ to the nearest point of the form $k/LCM(n,m)$, where $k\in Z$. To maximize the function $f(t)$, the value of $t$ should be chosen so that it is exactly in the middle between the neighboring points $k/LCM(n,m)$ and $(k+1)/LCM(n,m)$, where $k$ is an integer. The value of the function at such a point is $1/2\cdot LCM(n,m)$, which will be the answer.

Example

Let $f(t)=mi{n}_{i,j\in Z}|\frac{i}{3}-\frac{j}{5}+t|$. It is obvious that $f(0)=0$. The equality $f(1/15)=0$ holds because there exist integers $i$ and $j$ such that $i/3-j/5=-1/15$ (for example, $i=1,j=2$). From this, it also follows that $f(k/15)=0$, where $k\in Z$.

However, $f(k/15+1/30)=1/30$, and this will be the maximum value of the function, as the point $k/15+1/30$ is at the maximum distance from the zeros of the function.

Algorithm realization

The gcd function computes the greatest common divisor of the numbers $a$ and $b$.

long long gcd(long long a, long long b)
{
  return b == 0 ? a : gcd(b, a % b);
}

The lcm function computes the least common multiple of the numbers $a$ and $b$.

long long lcm(long long a, long long b)
{
  return a / gcd(a, b) * b;
}

The main part of the program. Read the input data.

while (scanf("%lld %lld", &n, &m) == 2)
{

Compute and print the result.

  d = 2 * lcm(n, m);
  printf("%lld/%lld\n", 1, d);
}