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
2021-04-26 17:16:29
how to check the code
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
2021-03-14 16:28:02
time limit exceeded, what is it? and what might be the issue here?
2021-02-11 15:53:23
If you got WA...then try for n=1 *_*
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
2020-12-01 16:38:04
0.33ms use primefactorization trick
2020-11-02 16:47:41
got wrong for \n .. lol
2020-10-23 10:20:49
AC in one go
2020-10-14 19:33:04
AC in one go! Time: 0.07
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.
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