uva10976

maksyuki 发表于 oj 分类，标签:
11 6月 2018
0

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

 

题目类型：暴力枚举

算法分析：直接枚举出y，然后再计算出x是否满足要求，由x >= y，则可得1/x <= 1/y，代入表达式可得y <= 2k，所以枚举[1,2k]内的值作为y，然后生成x并判断