分数拆分（Fractions Again?!, UVa 10976）

题目链接：https://vjudge.net/problem/UVA-10976

It is easy to see that for every fraction in the form 1k(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

题意概述：输入正整数k，找到所有的正整数x≥y，使得1/k=1/x+1/y;

解题思路：既然要求找出所有的x、y，枚举对象自然就是x、y了。可问题在于，枚举的范围如何？ 从1/12=1/156+1/13可以看出，x可以比y大很多。难道要无休止地枚举下去？当然不是。因为1/k=1/x+1/y，所以1/k>1/y,所以y>k;又因为x>=y,所以1/x≤1/y，且1/k=1/x+1/y，所以1/k-1/y≤1/y，得y≤2k；这样，只需要在k+1到2k范围之内枚举y，然后根据y尝试 计算出x即可，因为1/k=1/x+1/y，得x=ky/(y-k)。暴力枚举也需要进行数学推倒可以大大缩小枚举的范围。

代码如下：

#include<iostream>
using namespace std;
const int maxn=1000;
int ansx[maxn],ansy[maxn];
int main()
{
    int k;
    int cnt;
    while(cin>>k)
    {
        cnt=0;
        for(int j=k+1;j<=2*k;j++)
        {
            if((k*j)%(j-k)==0&&k*j/(j-k)>=j)
            {
                ansx[cnt]=k*j/(j-k);
                ansy[cnt]=j;
                cnt++;
            }
        }
        cout<<cnt<<endl;
        for(int i=0;i<cnt;i++)
        {
            cout<<"1/"<<k<<"=1/"<<ansx[i]<<"+1/"<<ansy[i]<<endl;
        }
    }
    return 0;
}