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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
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
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2017-06-03 05:36:50
:) happy AC Last edit: 2017-06-03 05:37:41 |
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2017-05-24 17:57:13
Instead of thinking for which number will loop stop,think for which situation the loop will stop. |
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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. |
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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 |
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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. |
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2017-04-03 19:50:32
HINT: If the number is a power of 2 then it will stop otherwise not. |
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2017-03-15 08:41:02
use unsigned long long int , long long int costed me WA |
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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. |
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2017-01-29 17:42:45
No fancy things required, just try to find out pattern for few starting numbers using pen and paper |
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2016-12-27 14:54:26 ~~diva~~
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 |