| Submit | All submissions | Best solutions | Back to list |
FCDC - Factorial Modulo |
You are given 2 integers a, b. Find the number of i for which i! is divisible by a but not b. if i! is divisible by a and b, then you should not count that i.
Input
One line that contains a and b.
Output
Output the result in one line.
Example
Input: 2 3 Output: 1
Constraints
1 ≤ a ≤ b ≤ 107
Explanation
2! is the only factorial which is divisible by 2 and not divisible by 3.
| Added by: | Ruhan Habib |
| Date: | 2015-11-05 |
| Time limit: | 1s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: ASM64 GOSU JS-MONKEY |
| Resource: | Own Problem |
hide comments
|
||||||||
|
2021-09-27 03:55:30
your answer should not be negative |
||||||||
|
2021-06-26 11:23:40
Factorization + Binary search Loved the problem :D |
||||||||
|
2019-02-09 16:40:31 tanardi gunawan
good problem |
||||||||
|
2018-11-22 16:29:22
nice problem..enjoyed solving it.Thanks Ruhan! |
||||||||
|
2018-08-20 09:10:30
did it using binary search and highest power of a prime in n!. Enjoyed the problem. Hope it helps someone in need. All the best!!! |
||||||||
|
2018-07-01 17:06:34
AC in 10 min :) Last edit: 2018-07-01 17:07:43 |
||||||||
|
2018-05-26 11:46:44
Factorization and brain storming :) |
||||||||
|
2018-05-18 22:43:37
that's a easy one :) factorization+simple logic |
||||||||
|
2017-10-24 16:53:15
dont think too much,,simple logic will get you through |
||||||||
|
2017-10-20 06:04:48
Seemed easy but took me good 2 hours to crack. The implementation is not so straightforward either - my code came out quite bloated - but considering half of C submissions are slower than my Python3 one, you can probably bruteforce in a few lines also. |
||||||||
RSS