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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visvats_141095:
2016-06-05 19:08:56
AC in first go! :D |
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azam_9:
2016-06-03 08:59:07
modified sieve also works.. |
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katyalch:
2016-05-16 05:48:30
uh oh tle |
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manav_itmu:
2016-05-15 21:32:31
tle. any help ? |
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shubhanshu001:
2016-05-15 15:35:05
AC in first go!!..:) |
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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) |
| 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 |
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