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
hide comments
Santiago Palacio:
2013-04-15 15:57:39
@ alg0: an O(sqrt(n)) algorithm should pass, mine did, look for I/O optimizations. Last edit: 2011-06-13 06:07:00 |
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SD:
2013-04-15 15:57:39
Why TLE??? Although i m generating the array for sum only one time??? |
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Ocean Blue:
2013-04-15 15:57:39
do you have to use buffers to avoid TLE?? |
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alg0_seekar:
2013-04-15 15:57:39
hi guys i m using O(sqrt(n) algorithm still getting TLE..can any one suggest what optimization still i needed..?? |
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Piotr KÄ…kol:
2013-04-15 15:57:39
@Neeraj Bhat - Yup. |
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Neeraj Bhat:
2013-04-15 15:57:39
is output for 500000 is 730453 ??? |
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bashrc is back:
2013-04-15 15:57:39
precomputation+some mathematics=AC |
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Santiago Palacio:
2013-04-15 15:57:39
TLE? i dont believe it! i think i'm using a good algorithm, and efficient io. |
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!!!!!:
2013-04-15 15:57:39
it depends how you factorize ur number..... other wise u'll get tle |
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sri:
2013-04-15 15:57:39
why TLE??? i was using a nice loopings to calculate.. still tle???? |
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 |