DIVSUM - Divisor Summation
Given a natural number n (1 ≤ n ≤ 500000), please output the summation of all its proper divisors.
Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.
e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.
Input
An integer stating the number of test cases (equal to about 200000), and that many lines follow, each containing one integer between 1 and 500000 inclusive.
Output
One integer each line: the divisor summation of the integer given respectively.
Example
Sample Input: 3 2 10 20 Sample Output: 1 8 22
Warning: large Input/Output data, be careful with certain languages
hide comments
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manav_itmu:
2016-05-15 21:32:31
tle. any help ? |
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mohitgupta07:
2016-05-08 21:12:56
yeah...simple maths..:) :) feeling gud after doing it wid just 1 try(actually got 1 wa coz of that n=1 )..yeah :D |
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shanti_chai:
2016-04-24 15:43:14
@hrithik,vijay... calculate divisors up to sqrt(n) . For better runtime you can use modified sieve |
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hrithik_sethia:
2016-04-19 19:59:31
getting tle please how do we do this?? |
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vijayreddie:
2016-03-24 06:12:37
time limit exceeded how to |
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pranjalikumar9:
2016-02-14 18:27:14
Just need to know that divisors exist in pairs. If i divides n then n/i also divides n. And smaller of the 2 divisors lies below sqrt(n) |
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supratik101:
2016-02-10 18:10:03
How to wite test case in program?
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Abhishek Kumar Singh:
2016-02-01 08:20:43
For those who are getting WA, take care of the square no.'s.That is no's such as 16. |
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Nallagatla Manikanta:
2016-01-19 16:25:07
For n=1, answer should be 0. Cost me 1 WA!! Got AC finally! |
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shohjahon:
2016-01-17 19:09:07
Hello everybody, Could anybody explain me why runtime error is occuring in my judge, (I have tested the program on my c++ compiler , the answer is being correcly for every case also 1) |
Added by: | Neal Zane |
Date: | 2004-06-10 |
Time limit: | 3s |
Source limit: | 5000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All |
Resource: | Neal Zane |