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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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2021-04-26 17:16:29
how to check the code |
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2021-04-18 22:15:48
Solve it using O(Sqrt(n)) AC in one go : 0.46ms C++ Solution(Only if you have tried) : <snip> Last edit: 2023-03-08 17:53:45 |
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2021-03-14 16:28:02
time limit exceeded, what is it? and what might be the issue here? |
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2021-02-11 15:53:23
If you got WA...then try for n=1 *_* |
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2021-02-08 13:53:55
Getting runtime error don't know why please help https://www.spoj.com/submit/DIVSUM/id=27407764 -- code link written in java Got it my bad.. Last edit: 2021-02-08 14:01:56 |
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2020-12-01 16:38:04
0.33ms use primefactorization trick |
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2020-11-02 16:47:41
got wrong for \n .. lol |
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2020-10-23 10:20:49
AC in one go |
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2020-10-14 19:33:04
AC in one go! Time: 0.07 |
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2020-10-07 13:41:44
We cannot use i<=sqrt(num). if we use num=10 then sqrt(10)<4. So the answer will be 3 which is incorrect. |