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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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2020-08-02 11:23:37
Ac in one go |
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2020-07-18 17:23:26
AC. Time: 0.50ms. Lang.: C++ Used: sieve of eratosthenes, binary exponentiation, sum of divisors formula (after prime factorization). |
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2020-07-15 17:35:36
At least AC. After eating 6 WA. LOL. 0.74 MS |
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2020-07-13 00:57:34
say the wrong answer but I don't understand why |
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2020-06-16 05:00:55
Python users getting TLE should use stdin and stdout |
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2020-05-28 13:00:53
why it give run time error |
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2020-05-18 13:09:46
AC in one go!! |
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2020-05-16 08:47:50
AC IN ONE GO |
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2020-05-12 10:50:11
getting right answers in my system(Just according to formula) still getting wrong ans here... what else should I take care of? |
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2020-04-24 12:03:20
use i<=sqrt(num) to avoid TLE |