SPOJ.com - Problem WILLITST

WILLITST - Will it ever stop

#ad-hoc-1

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 !

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

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

Input:
4

Output:
TAK

< Previous 4 5 6 7 8 9 10 11 12 13 14 Next >

	wannacry123: 2017-06-11 16:41:09 Using unsigned long long int is the key Even long long gives the wrong ans
	yank: 2017-06-03 05:36:50 :) happy AC Last edit: 2017-06-03 05:37:41
	senacoder111: 2017-05-24 17:57:13 Instead of thinking for which number will loop stop,think for which situation the loop will stop.
	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