ABA12D - Sum of divisors!
If you really want to learn something by solving this problem, don't hard code!
There is a nice logic behind this!
Kartheeswaran was recently reading an article on perfect numbers, whose sum of divisors equals twice the number. He was intrigued by them and decided to generate them but to his disappointment they turned out to be quite rare. So he decided to look out for a different property related to sum of divisors. What is more interesting than a number being a prime? So he decided to look out for numbers whose sum of divisors is a prime number and he was the inventor of these special numbers he gave them the name K-numbers.
Given a range [A,B] you are expected to find the number of K-numbers in this range.
The first line of input indicates the number of test cases T.
Then in the following T lines there will be a pair of integers A and B.
Output T lines each containing a single integer ‘c’ which denotes the number of K-numbers which lie in the interval [A,B] inclusive of the end points.
Explanation of sample case:
1) In the range [1,5] the K-numbers are 2 and 4 because
Divisors of 2 are 1 and 2 which sum up to 3, which is a prime.
Divisors of 4 are 1,2 and 4 which sum up to 7, which is a prime.
2) The K-number in the range [9,10] is 9.
nice logic :)
My 100th ^..^
It's really nice :)
I factorized only the numbers which are perfect squares. There has to be a better way to solve this..
easy one...AC in one go ;-)
Implemented brute force approach but got TLE , eventually hard coded . ;((
Rather than say "If you really want to learn something by solving this problem, don't hard code!", you might as well have increased the time limit to ~3s and the query limit to 10^12; that would have made hard-coding without the proper solution impractical.
can take help from
|Added by:||Kashyap Krishnakumar|
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