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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 |
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2017-12-30 14:53:54
https://mathschallenge.net/library/number/sum_of_divisors...... this could help!! |
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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 |
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2017-12-19 13:13:15
why i am getting time limit exceeded?? can someone please help |
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2017-12-16 13:54:55
remember: sqrt(x)*sqrt(x)!=x vs (int)sqrt(x)*(int)sqrt(x)==x |
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2017-12-06 14:42:31
Somebody solve this problem using python? |
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2017-12-03 08:35:43
AC?? Help What is this? _/\_ |
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2017-12-01 12:27:13
can anyone help me understand hoow to use stdin.read() in python |
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2017-11-06 04:14:17
why are we taking square root of n ? |
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2017-08-12 14:33:06
AC in one Go....! simple ....... |
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2017-07-26 19:53:38
easy one solved in 0.06 s |