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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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2017-07-01 21:49:44
take care of perfect squares |
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2017-06-30 15:41:27
for those getting WA some testcases: 1 -> 0 99999 -> 48513 500000 -> 730453 9689 -> 1 12345 -> 7431 |
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2017-06-23 08:18:24
it is showing TLE in C please help |
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2017-06-20 01:07:09
Answer fits in long long for C++. |
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2017-06-19 01:31:22
what are the test cases i am getting wrong answer? |
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2017-05-31 05:29:48
Guys I am getting TLE in python I don't know why please help |
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2017-05-29 19:49:34
Guys take care of n = 1, the solution is 0, cost me 1 WA! |
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2017-04-29 00:28:23
<1. Don't post any source code here. > i dont know why judge is telling my answer wrong anybody help Last edit: 2017-04-29 05:57:29 |
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2017-04-19 22:04:01
for num=1, ans =0, for perfect square count divisor once only |
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2017-04-03 13:37:13
Simple Problem..!! take care for n=1 ans should be 0 Cost me 2 WA's..!! |