Uva - 10976 - Fractions Again?!

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

212

Sample Output

21/2 = 1/6 + 1/31/2 = 1/4 + 1/481/12 = 1/156 + 1/131/12 = 1/84 + 1/141/12 = 1/60 + 1/151/12 = 1/48 + 1/161/12 = 1/36 + 1/181/12 = 1/30 + 1/201/12 = 1/28 + 1/211/12 = 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\n", cnt);for (int i = 0; i < cnt; i++) {printf("1/%d = 1/%d + 1/%d\n", k, ans.front().first, ans.front().second);ans.pop(); // 全部输入就刚好清空了}}return 0;}