Hailstone Numbers

View as PDF

Submit solution

All submissions

Best submissions

Points: 3

Time limit: 2.0s

Memory limit: 64M

Author:

WhiteTofuLord

Problem type

Pick a positive integer $n$ ~n~. If it is odd, multiply it by three and then add one. If $n$ ~n~ is even, divide it by two. The positive integer obtained is the new $n$ ~n~, and this is repeated until the number becomes 1. Given the value of $n$ ~n~, with $1 \le n< 2^{31}$ ~1 \le n< 2^{31}~, determine the number of operations before $n$ ~n~ becomes 1.

Sample Input

Sample Output

Explanation

$n$ ~n~ will become 10, 5, 16, 8, 4, 2, then 1, which is a total of 7 operations.

Comments

8

jason6 commented on Oct. 27, 2018, 10:06 p.m.

Collatz Conjecture!

2

SeanJxie commented on Dec. 18, 2019, 3:04 p.m.

Someone has been watching NumberPhile!

4
aeternalis1 commented on Nov. 18, 2017, 3:14 p.m. ← edit 2 →
For the full explanation:

Stop asking for help in comments.

Go to https://discord.com/invite/EgJVpxz and ask for help in the #help channel.