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
2017-12-30 14:53:54
https://mathschallenge.net/library/number/sum_of_divisors...... this could help!!
2017-12-28 09:41:25
Use Sieve for pre computation and solve each query in O(1) .Dont Forget to use long. you'll surely endup with fastest submission :) 0.00s

Last edit: 2017-12-28 09:54:07
2017-12-19 13:13:15
why i am getting time limit exceeded??
can someone please help
2017-12-16 13:54:55
remember: sqrt(x)*sqrt(x)!=x vs (int)sqrt(x)*(int)sqrt(x)==x
2017-12-06 14:42:31
Somebody solve this problem using python?
2017-12-03 08:35:43
AC?? Help What is this?
_/\_
2017-12-01 12:27:13
can anyone help me understand hoow to use stdin.read() in python
2017-11-06 04:14:17
why are we taking square root of n ?
2017-08-12 14:33:06
AC in one Go....! simple .......
2017-07-26 19:53:38
easy one solved in 0.06 s
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