Uva

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 andy 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

遍历y，y的取值是从k+1到2k，每次判断 y * k % (y - k)是否是0，就是判断k之分一减去y分之一能否约分成分子为1的分数，结果用队列存放。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <bitset> 
#include <cassert> 

using namespace std;

queue<pair<int, int> > ans;

int main()
{
	int k;
	while (cin >> k && k) {
		int cnt = 0;
		for (int i = k + 1; i <= 2 * k; i++) {
			long long pro = (long long)i * k;
			int minus = i - k;
			if (pro % minus == 0) { // 是否能约分成分子是1的分数
				ans.push(pair<int, int>(pro / minus, i));
				cnt++;
			}
		}
		printf("%d
", cnt);
		for (int i = 0; i < cnt; i++) {
			printf("1/%d = 1/%d + 1/%d
", k, ans.front().first, ans.front().second);
			ans.pop(); // 全部输入就刚好清空了
		}
	}

	return 0;
}