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

Help him finding it out !


In first line one number n<=10^14.


Print "TAK" if program will stop, otherwise print "NIE"




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ayushgupta1997: 2017-05-20 06:52:18

ac in one go! simply try to figure out from 1 to 10 using pen and no need of unsigned long long and bit manipulation,while loop will do ur job.

newbie_127: 2017-05-17 15:45:07

AC in one go .
My approach : use a counter (10^8)
Do what the code says . Increment counter every time you iterate.
If counter exceeds (10^8) , print NIE , else print TAK .
Weak test cases I would say :P

bharat190991: 2017-04-27 05:49:05

if the loop is infinite then it will repeat the pattern so no need of any bit manipulation. hope it will help.

horcrux2301: 2017-04-03 19:50:32

HINT: If the number is a power of 2 then it will stop otherwise not.

rohit659: 2017-03-15 08:41:02

use unsigned long long int , long long int costed me WA

rohit9934: 2017-02-11 15:51:41

Bit manipulation Says that if( x&(x-1))==0 then number is a power of 2,else not.Dont forget the brackets,it costs me a WA.

bhaskar_uoh: 2017-01-29 17:42:45

No fancy things required, just try to find out pattern for few starting numbers using pen and paper

~~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
deepak1228: 2016-12-21 18:11:49

wanna learn something new from this question go through above link :)
in o(1) time......ac

apurvgs: 2016-12-20 14:41:20

easy problem,,just if condition and bit manipulation

Last edit: 2016-12-24 13:27:44

Added by:Krzysztof Lewko
Time limit:0.906s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64
Resource:AMPPZ 2011