UVa 10976 - Fractions Again?!

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

容易看出y的范围是(k , 2k]，x大于等于2k，但上界不确定
因此可看做k和y已知，求是1/k - 1/y结果的分子是否为1
通分后即求k*y是否可被y-k整除

代码如下

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int maxn=100000;int main(){    int n;    while(cin>>n){        int i,j,k;        int x[maxn],y[maxn];        int cnt=0;        for(i=n+1 ; i<= n<<1 ; i++){            if((n*i)%(i-n) == 0){                x[cnt]=(n*i)/(i-n);                y[cnt]=i;                cnt++;            }        }        cout<<cnt<<endl;        for(i=0 ; i<cnt ; i++)            printf("1/%d = 1/%d + 1/%d\n",n,x[i],y[i]);    }    return 0;}