IITD4 - Divisor Summation Powered
Define F(n,k) = Sum of kth powers of all divisors of n
So for example F(6,2) = 1^2 + 2^2 + 3^2 + 6^2 = 50
Define further G(a,b,k) as : Sum of F(j,k) where j varies from a to b both inclusive
Your task is to find G(a,b,k) given a,b & k.
As values of G can get very large , you only need to output the value of G(a,b,k) modulo 10^9+7.
First line of input file contains a single integer T - denoting the number of test cases.
The follow description of T test cases. Each test case occupies exactly one line which contains three space separated integers a,b & k.
Output your result for each test case in a new line.
Sample Input File:
2 2 1
1 3 2
Sample Output File:
Description of sample output:
In case 1, we are to find sum of divisors of 2. which is nothing but 1+2=3.
In case 2, we are to find sum of squares of divisors of 1 2 & 3. So for 1 sum is = 1. For 2 sum is = 1^2+ 2^2= 5. For 3 sum is = 1^2 + 3^2=10.
So ans is 16.
Number of test cases <=20
Learnt some new things, use modular exponentiation for faster power calculation
easy one just simple logic is enough :)
consider ** times=b/i - (a-1)/i ** caused me two WA..
O(sqrt(b)+sqrt(a))logn giving AC in 15 Sec .. :(
Got AC with a generic implementation.
oye lakshman help plz:
@lakshman is this the correct ans 299384888 also is there any fast algo for powersum
@crazzysuarez Have you tried this case
getting WA plz provide with some test case
yay :DDDDDD finalyy AAAcCCCCC :)
nlog(n) always giving TLE on running (10)