UVA - 10976

Problem A: Fractions Again?!

Time limit: 1 second

It is easy to see that for every fraction in the form  (k > 0), we can always find two positive integers x and y, x ≥ y, such that: 

.

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 <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

char ch[1000][50];
int main () {
    int n;
    while (cin >> n) {
        int pos = 0;
        for (int i = n + 1;i <= 2 * n;i++) {
            if (i * n % (i - n) == 0) {
                sprintf(ch[pos++],"1/%d = 1/%d + 1/%d\n",n,i * n / (i - n),i);
            }
        }
        cout << pos << endl;
        for (int i = 0;i < pos;i++) {
            cout << ch[i];
        }
    }
}