Fractions Again?! UVA – 10976

It is easy to see that for every fraction in the form 1 k (k > 0), we can always find two positive integers x and y, x ≥ y, such that: 1 k = 1 x + 1 y Now our question is: can you write a program that counts how many such pairs of x and y there are for any given k?
Input Input contains no more than 100 lines, each giving a value of k (0 < k ≤ 10000).
Output
For each k, output the number of corresponding (x,y) pairs, followed by a sorted list of the values of x and y, as shown in the sample output.
Sample Input
2 12
Sample Output
2
1/2 = 1/6 + 1/3
1/2 = 1/4 + 1/4
8
1/12 = 1/156 + 1/13
1/12 = 1/84 + 1/14
1/12 = 1/60 + 1/15
1/12 = 1/48 + 1/16
1/12 = 1/36 + 1/18
1/12 = 1/30 + 1/20
1/12 = 1/28 + 1/21
1/12 = 1/24 + 1/24

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>using namespace std;int main(int argc, const char * argv[]){int k, x, y;while(cin >> k){int cnt = 0;for(y = k + 1; y <= 2 * k; y++){if((k * y) % (y - k) == 0){x = (k * y) / (y - k);printf(\"1/%d = 1/%d + 1/%d\\n\", k, x, y);cnt++;}}cout << cnt <<endl;}return 0;}