blob: 038e44ab2ada6b056d46b6f7ef968bf77fb630cb (
plain) (
blame)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
|
# [[2016-10-03] Challenge #286 [Easy] Reverse Factorial](https://www.reddit.com/r/dailyprogrammer/comments/55nior/20161003_challenge_286_easy_reverse_factorial/)
## Description
Nearly everyone is familiar with the factorial operator in math. 5! yields 120
because factorial means "multiply successive terms where each are one less than
the previous":
5! -> 5 * 4 * 3 * 2 * 1 -> 120
Simple enough.
Now let's reverse it. Could you write a function that tells us that "120" is
"5!"?
Hint: The strategy is pretty straightforward, just divide the term by
successively larger terms until you get to "1" as the resultant:
120 -> 120/2 -> 60/3 -> 20/4 -> 5/5 -> 1 => 5!
## Sample Input
You'll be given a single integer, one per line. Examples:
120
150
## Sample Output
Your program should report what each number is as a factorial, or "NONE" if
it's not legitimately a factorial. Examples:
120 = 5!
150 NONE
## Challenge Input
3628800
479001600
6
18
## Challenge Output
3628800 = 10!
479001600 = 12!
6 = 3!
18 NONE
|
