UVA - 10976 Fractions Again?!

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

注意找k,x,y之间的规律和范围，不能盲目暴力啊

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int k;
    while(scanf("%d", &k)!=EOF)
    {
        int x, y, cnt = 0;
        vector<int> res1, res2;
        for(y = k+1; y <= 2*k; y++)
        {
            x = (k*y)/(y-k);
            if(k*y == (y-k)*x){
                cnt++;
                res1.push_back(x);
                res2.push_back(y);
            }
        }
        printf("%d\n", cnt);
        for(int i = 0; i < cnt; i++)
            printf("1/%d = 1/%d + 1/%d\n", k, res1[i], res2[i]);
    }
    return 0;
}