uva-10976 - Fractions Again

Problem A: Fractions Again?!

Time limit: 1 second

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 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).

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.

2 12

1/2 = 1/6 + 1/3

1/2 = 1/4 + 1/4

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

Problemsetter: Mak Yan Kei

解題策略

分析題目

1/k <= 1/y+1/y=2/y => y<=2k 且 y>k

列舉y由k+1到2k

1/k=1/x+1/y=(x+y)/xy => kx+ky=xy => ky=x(y-k) => x=ky/(y-k)

y=k+1與y=2k一定是解