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
(Tjandra Satria Gunawan)(曾毅昆):
2013-04-15 15:57:39
my best time is 0.13s in C++
|
|
Francky:
2013-04-15 15:57:39
I've work many times on this problem since almost a year. This is THE problem that push me to learn C. I haven't the fastest chrono, but I'm very happy today with this one. \o/
|
|
xyb:
2013-04-15 15:57:39
Just replace cin/cout with scanf/printf, got AC. |
|
rb:
2013-04-15 15:57:39
finally got AC!!!!!!! |
|
useless:
2013-04-15 15:57:39
Super easy..move to tutorials.. |
|
srishti goel:
2013-04-15 15:57:39
my code is working in ideone bt showing a run time error here..
|
|
eliminator:
2013-04-15 15:57:39
@secret22
|
|
secret22:
2013-04-15 15:57:39
tle... :( help me....... |
|
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?
|
|
nikoo28:
2013-04-15 15:57:39
imp test case:
|
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 |