CCC '96 S1 - Deficient, Perfect, and Abundant

Points: 5

Time limit: 2.0s

Memory limit: 64M

Problem type

Write a program that repeatedly reads a positive integer, determines if the integer is deficient, perfect, or abundant, and outputs the number along with its classification.

A positive integer, $n$ ~n~, is said to be perfect if the sum of its proper divisors equals the number itself. (Proper divisors include $1$ ~1~ but not the number itself.) If this sum is less than $n$ ~n~, the number is deficient, and if the sum is greater than $n$ ~n~, the number is abundant.

The input starts with the number of integers that follow. For each of the following integers, your program should output the classification, as given below. You may assume that the input integers are greater than $1$ ~1~ and less than $32\,500$ ~32\,500~.

Sample Input

Sample Output

4 is a deficient number.
6 is a perfect number.
12 is an abundant number.

CCC problem statements in large part from the PEG OJ

Comments

0

ayay9 commented on May 26, 2021, 12:58 p.m.

THinking of Generating all combos but then i think it'll TLE

5

QWERTYL commented on Dec. 13, 2020, 6:35 p.m.

It took me a long time to realize I was outputting "a abundant" instead of "an abundant"

-1

myl commented on Feb. 20, 2021, 7:48 p.m.

Same mistake here too

0

Dav_Did commented on Feb. 16, 2021, 2:56 p.m.

Thanks, that help me out because I fall for it too

0

Nate commented on Oct. 29, 2020, 11:22 a.m.

I have a strange comment, kind of related to the question, how many perfect integers are there up until say up until 2,147,483,647? I wrote a program, but in about 15 minutes it only got 4. And those where 6, 28, 496, and 8128. Has anybody else been able to actually get a complete list?

8
d commented on Oct. 29, 2020, 2:02 p.m.
A quick search found a good answer to your question. In case you're too lazy to read the link, I can summarize:

If $2^p-1$ ~2^p-1~ is prime, then $2^{p-1} (2^p-1)$ ~2^{p-1} (2^p-1)~ is a perfect number. The first few values of $p$ ~p~ are 2, 3, 5, 7, 13 and their corresponding values are 6, 28, 496, 8128, 33550336, ...

All other even numbers are not perfect.

The current limitations are:

Right now, there are 51 known values of $p$ ~p~ (the largest is 82589933) and 51 known perfect numbers. We don't know if there are finitely or infinitely many $p$ ~p~.

It is unknown if any odd numbers are perfect. An odd perfect number must be greater than $10^{1500}$ ~10^{1500}~.

0

feliser commented on Oct. 29, 2020, 1:40 p.m.

https://en.wikipedia.org/wiki/Perfect_number

7

CountT commented on Sept. 10, 2020, 5:00 p.m.

Forgot that you have to use an instead of a before another word that starts with a vowel. FP -> Facepalm
(－‸ლ)

0

chessdongdong commented on Feb. 1, 2020, 11:11 a.m.

Forgot the period, whoops

-4

SeanJxie commented on Nov. 11, 2019, 12:49 p.m.

My code acts differently when running in the judge?

-5

Narcariel commented on Oct. 23, 2019, 8:48 p.m. ← edited →

Hey does anyone know why I am getting an error on the first test case? I dont know cause im a bot.

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1

Dingledooper commented on Oct. 23, 2019, 9:34 p.m.

In your for-loop, is there a reason you are looping up until num/2+2? It doesn't seem intuitive why you need to add 2.

-21

WEAVER commented on Feb. 11, 2019, 6:53 p.m. ← edited →

WAY TOO HARD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! NOT EVEN CLEAR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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3

Arihan10 commented on Feb. 13, 2019, 9:34 a.m. ← edit 2 →

What do you not understand? Perhaps your comment could be better phrased simply stating your problem.

-7

rsylshzxdkh commented on June 21, 2017, 4:22 p.m.

1 + 2 + 3 = 6 and the question says the sum does not include the number itself?

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3

DKLS2 commented on June 21, 2017, 7:55 p.m.

Because it is not clear, I am going to assume you are asking a question.

Yes, the question says the sum does not include the number itself as every number would be abundant (for numbers larger than 1).