WILLITST - Will it ever stop
When Bob was in library in University of Warsaw he saw on one of facades caption :"Will it ever stop?" and below some mysterious code:
while n > 1
if n mod 2 = 0 then
n:=n/2
else
n:=3*n+3
Help him finding it out !
Input
In first line one number n<=10^14.
Output
Print "TAK" if program will stop, otherwise print "NIE"
Example
Input: 4 Output: TAK
hide comments
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~~diva~~:
2016-12-27 14:54:26
no need of bit maipulation , just check if the number is becoming 3,6 or 12 at any time within the while loop .if yes, only then it's forming an infinite loop! Last edit: 2016-12-27 15:07:49 |
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deepak1228:
2016-12-21 18:11:49
https://www.hackerearth.com/practice/notes/bit-manipulation/
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apurvgs:
2016-12-20 14:41:20
easy problem,,just if condition and bit manipulation Last edit: 2016-12-24 13:27:44 |
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scorpion_ajay:
2016-12-19 17:02:38
use your pen :p |
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humble_fool:
2016-12-09 07:53:15
AC in one go. Bits manipulation is the key. Although you can do it using log2 but try Bits Manipulation, you will learn something new. :) |
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spartax:
2016-11-30 07:07:47
HINT: if n is a power of 2, it gets reduced to 1. and if n is not a power of 2, then if n is even, it becomes odd. Think what happens to odd integers and you'll get theta(1) solution :-) |
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radeonguy:
2016-11-28 18:29:04
HINT : you may want to use double instead of float if using log2 |
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mohitgupta07:
2016-11-17 10:21:02
Donot use that log function. It will give you wa... find another way to get power of 2. umm... how about bit manipulation ;) |
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aditya_rev:
2016-11-03 18:01:28
ez as brainf##ck |
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sudiptob2:
2016-10-24 09:12:06
For those who are getting WA for log(n)/log(2)
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| Added by: | Krzysztof Lewko |
| Date: | 2011-11-09 |
| Time limit: | 0.906s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: ASM64 |
| Resource: | AMPPZ 2011 |
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