Given an integer N. The task is to find the maximum possible sum of intermediate values (Including N and 1) attained after applying the beow operation:
Divide N by any divisor (>1) until it becomes 1.
Input: N = 10 Output: 16 Initially, N=10 1st Division -> N = 10/2 = 5 2nd Division -> N= 5/5 = 1 Input: N = 8 Output: 15 Initially, N=8 1st Division -> N = 8/2 = 4 2nd Division -> N= 4/2 = 2 3rd Division -> N= 2/2 = 1
Approach: Since the task is to maximize the sum of values after each step, try to maximize individual values. So, reduce the value of N by as little as possible. To achieve that, we divide N by its smallest divisor.
Below is the implementation of the above approach:
Time Complexity: O(sqrt(n)*log(n))
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