Fractions Again?! UVA

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 <iostream>//求1/n = 1/x + 1/y

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <string>

using namespace std;

int x[],y[];

int main()

{

    int n,cnt;

    while(cin >> n)

    {

        memset(x,,sizeof(x));

        memset(y,,sizeof(y));

        cnt = ;

        for(int i=n+;i<=*n;i++)

        {

            if((i*n)%(i-n)==)//注意这个判断条件，判断等式是否可能成立

            {

                x[cnt] = (i*n)/(i-n);//求x的式子

                y[cnt] = i;

                cnt++;

            }

        }

        printf("%d\n",cnt);

        for(int i=;i<cnt;i++)

            printf("1/%d = 1/%d + 1/%d\n",n,x[i],y[i]);

    }

    return ;

}

巴特西

Fractions Again?! UVA - 10976

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