Sum of the largest odd divisors of the first n numbers - algorithm

Question

I've been working on topcoder recently and I stumbled upon this question which I can't quite make understand.
The question is to find F(n) = f(1)+f(2)+....+f(n) for a given "n" such that f(n) is the largest odd divisor for n.
There are many trivial solutions for the answer; however, I found this solution very intriguing.
int compute(n) {
if(n==0) return 0;
long k = (n+1)/2;
return k*k + compute(n/2);
}
However, I don't quite understand how to obtain a recursive relation from a problem statement such as this. Could someone help out?

Answer 1

I believe they are trying to use the following facts:
f(2k+1) = 2k+1, i.e. the largest odd divisor of an odd number is the number itself.
f(2k) = f(k). i.e the largest odd divisor of an even number 2m is same as the largest odd divisor of the number m.
Sum of first k odd numbers is equal to k^2.
Now split {1,2,..., 2m+1} as {1,3,5,7,...} and {2,4,6,...,2m} and try to apply the above facts.

Answer 2

You can use dynamic approach also using auxiliary spaces
int sum=0;
int a[n+1];
for(int i=1;i<=n;i++){
if(i%2!=0)
a[i] = i;
else
a[i] = a[i/2];
}
for(int i=1;i<=n;i++){
sum+=a[i];
}
cout<<sum;
As when number is odd then the number itself will be the greatest odd divisor and a[i] will store it's value and when number is even then the a[number/2] will be stored in a[i] because for even number the greatest odd divisor of number/2 will be the greatest odd divisor of the number.
It can also be solved using three cases when number is odd then add number itself else if number is power of 2 then add 1 else if number is even except power of 2 divide it by 2 till you get odd and add that odd to sum.

Answer 3

I cannot see how that algorithm could possible work for the problem you described. (I'm going to assume that "N" and "n" refer to the same variable).
Given n = 12.
The largest odd divisor is 3 (the others are 1, 2, 4, 6 & 12)
F(12) is therefor f(1) + f(2) + f(3) or 1 + 1 + 3 or 5.
Using this algorithm:
k = (12+1)/2 or 6
and we return 6 * 6 + f(6), or 36 + some number which is not going to be negative 31.

Answer 4

if this were Java, I'd say:
import java.util.*;
int sum_largest_odd_factors (int n){
ArrayList<Integer> array = new ArrayList();//poorly named, I know
array.add(1);
for(int j = 2; j <= n; j++){
array.add(greatestOddFactor(j));
}
int sum = 0;
for(int i = 0; i < array.size(); i++){
sum += array.get(i);
}
return sum;
}
int greatestOddFactor(int n){
int greatestOdd = 1;
for(int i = n-((n%2)+1); i >= 1; i-=2){
//i: starts at n if odd or n-1 if even
if(n%i == 0){
greatestOdd = i;
break;
//stop when reach first odd factor b/c it's the largest
}
}
return greatestOdd;
}
This is admittedly tedious and probably an O(n^2) operation, but will work every time. I'll leave it to you to translate to C++ as Java and J are the only languages I can work with (and even that on a low level). I'm curious as to what ingenious algorithms other people can come up with to make this much quicker.

Answer 5

IF u are looking for sum of all the odd divisors till n..
Sum of the all odd divisors of the first n numbers
...
for(long long int i=1;i<=r;i=i+2)
{
sum1=sum1+i*(r/i);
}
for sum of all divisors in a range l to r
for(long long int i=1;i<=r;i=i+2)
{
sum1=sum1+i*(r/i);
}
for(long long int i=1;i<l;i=i+2)
{
sum2=sum2+i*((l-1)/i);
}
ans=sum1-sum2;;;
THANK YOU!!