SPOJ.com - Problem WILLITST

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

Submit solution!

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

hide comments

2017-06-11 16:41:09
Using unsigned long long int is the key
Even long long gives the wrong ans

2017-06-03 05:36:50
:) happy AC

Last edit: 2017-06-03 05:37:41

2017-05-24 17:57:13
Instead of thinking for which number will loop stop,think for which situation the loop will stop.

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.

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

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.

2017-04-03 19:50:32
HINT: If the number is a power of 2 then it will stop otherwise not.

2017-03-15 08:41:02
use unsigned long long int , long long int costed me WA

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.

2017-01-29 17:42:45
No fancy things required, just try to find out pattern for few starting numbers using pen and paper