This is a very classical question. You are given a number in decimal respresentation and one of more decimal places in the tail of decimal part may be infinite repeated. You are asked to return the neareast fraction to that decimal. To be conventient, here we only considers decimal that is between 0.0 and 1.0. Let the number be denoted as , assume the last decimal places is repeating and we can see decimal place in total.

Assume , we have and , as are infinite repeating, so the decimal part of those two number are exactly the same, if we discount the second from the first, we will get:

To get the original , we can transform the equation into:

For the formal above always holds. So we can use GCD method to reduce the fraction on the right. As the question statement does not give us the exact value of , so we iterate from to and pick the one with lowest denominator as answer.

import java.io.*;
import java.util.*;

private static long gcd(long a, long b) {
if (b == 0) return a;
else return gcd(b, a % b);
}

public static void main(String[] args) throws IOException {
String[] str;
String s;
while ((s = in.readLine()) != null && !s.isEmpty()) {
if (s.equals("0")) break;

str = s.split("\\.");
String digits = str;
long n = Long.parseLong(digits);
int L = 1;
for(int i = 0; i < digits.length(); i++) L *= 10;

long a = 0, b = Long.MAX_VALUE;

for (int c = 10; c <= L; c *= 10) {
long u = n - n / c;
long d = (L / c) * (c - 1);

long g = gcd(u, d);

if (d / g < b) {
a = u / g;
b = d / g;
}
}

System.out.printf("%d/%d\n", a, b);
}

}
}