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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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2020-04-20 17:42:11
When I submit code in python3 for this problem, I get TLE but same logic when I write in cpp, it accepts. Why is this so? Last edit: 2020-04-20 18:51:41 |
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2020-01-21 17:26:26
for n=1 output is 0 |
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2019-08-08 12:24:56
corner case: n = 1 |
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2019-08-02 15:17:47
same solution in c++ give wrong answer and in c it is Ac |
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2019-07-27 13:54:01
did anyone know name of file input output please tell me |
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2019-07-02 06:11:34
Try DIVSUM2 :) |
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2019-04-21 19:56:37
if the number is 1 , the output must be 0. fml spent so much time on this stupid question!! |
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2019-04-15 08:19:39
it was an easy but remember answer of 1 is 0 always. |
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2019-04-13 09:56:13
AC in one shot use i*i<=num to avoid time isssue in this problem |
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2019-03-15 15:08:14
TOO easy, time limit should have been less |