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
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rb:
2013-04-15 15:57:39
finally got AC!!!!!!! |
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useless:
2013-04-15 15:57:39
Super easy..move to tutorials.. |
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srishti goel:
2013-04-15 15:57:39
my code is working in ideone bt showing a run time error here..
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eliminator:
2013-04-15 15:57:39
@secret22
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secret22:
2013-04-15 15:57:39
tle... :( help me....... |
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Shubham:
2013-04-15 15:57:39
O(t+m*m)(m is maxsize here i.e. 500000 and t,test cases) is faster than O(t*sqrt(n)). why is it so?
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nikoo28:
2013-04-15 15:57:39
imp test case:
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Monkey D. Luffy :
2013-04-15 15:57:39
For extra efficiency, remember that divisors come in pairs. For example, if 2 is a divisor of 20 then so is 20/2 = 10. |
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StupidGuy:
2013-04-15 15:57:39
Finally ACC..But still bad Time...:(
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Kartik Godawat:
2013-04-15 15:57:39
Please help. Tried every possible way for removing TLE. Computation done in <500ms for all inputs at local machine(every no <500000). Using Java BufferedReader. |
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 |