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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
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 |
hide comments
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2018-11-09 06:39:39
i get time limit exceed in python |
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2018-10-31 17:31:25
1 should be output 0 OMFG I have checked it for hours. |
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2018-09-24 17:40:51
AC in one go!!!!meow bheow kakkaw raar!!!! |
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2018-08-27 10:19:56
AC in ONE GO !!! Last edit: 2018-08-27 10:21:16 |
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2018-08-18 18:30:01
Do you have any test case? In PHP, I still get wrong answer. Please see my solution at: <snip> Last edit: 2023-03-08 17:56:03 |
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2018-05-12 17:41:48
Super easy waste of time ('_') |
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2018-04-19 06:30:30
why its time limit exceeded? |
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2018-03-29 09:02:46
the answer for 1 should be 0. It costs me two wrong answers. :( |
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2018-02-06 13:07:54
AC in ONE GO !!! Last edit: 2018-02-06 13:08:23 |
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2018-01-13 13:08:11
a nyc problem the simple logic is that we loop around for i*i<n think simple logic dont bother about learning sieve Last edit: 2018-01-13 13:09:03 |