Given a positive integer N, the task is to find the sum of divisors of first N natural numbers.
Examples:
Input: N = 4
Output: 15
Explanation:
Sum of divisors of 1 = (1)
Sum of divisors of 2 = (1+2)
Sum of divisors of 3 = (1+3)
Sum of divisors of 4 = (1+2+4)
Hence, total sum = 1 + (1+2) + (1+3) + (1+2+4) = 15Input: N = 5
Output: 21
Explanation:
Sum of divisors of 1 = (1)
Sum of divisors of 2 = (1+2)
Sum of divisors of 3 = (1+3)
Sum of divisors of 4 = (1+2+4)
Sum of divisors of 5 = (1+5)
Hence, total sum = (1) + (1+2) + (1+3) + (1+2+4) + (1+5) = 21
For linear time approach, refer to Sum of all divisors from 1 to N
Approach:
To optimize the approach in the post mentioned above, we need to look for a solution with logarithmic complexity. A number D is added multiple times in the final answer. Let us try to observe a pattern of repetitive addition.
Considering N = 12:
| D | Number of times added |
|---|---|
| 1 | 12 |
| 1 | 12 |
| 2 | 6 |
| 3 | 4 |
| 5, 6 | 2 |
| 7, 8, 9, 10, 11, 12 | 1 |
From the above pattern, observe that every number D is added (N / D) times. Also, there are multiple D that have same (N / D). Hence, we can conclude that for a given N, and a particular i, numbers from (N / (i + 1)) + 1 to (N / i) will be added i times.
Illustration:
- N = 12, i = 1
(N/(i+1))+1 = 6+1 = 7 and (N/i) = 12
All numbers will be 7, 8, 9, 10, 11, 12 and will be added 1 time only.- N = 12, i = 2
(N/(i+1))+1 = 4+1 = 5 and (N/i) = 6
All numbers will be 5, 6 and will be added 2 times.
Now, assume A = (N / (i + 1)), B = (N / i)
Sum of numbers from A + 1 to B = Sum of numbers from 1 to B – Sum of numbers from 1 to A
Also, instead of just incrementing i each time by 1, find next i like this, i = N/(N/(i+1))
Below is the implementation of the above approach:
Python3
# Python program for # the above approach mod = 1000000007 # Functions returns sum # of numbers from 1 to n def linearSum(n): return n*(n + 1)//2 % mod # Functions returns sum # of numbers from a+1 to b def rangeSum(b, a): return (linearSum(b) - ( linearSum(a))) % mod # Function returns total # sum of divisors def totalSum(n): # Stores total sum result = 0 i = 1 # Finding numbers and # its occurence while True: # Sum of product of each # number and its occurence result += rangeSum( n//i, n//(i + 1)) * ( i % mod) % mod; result %= mod; if i == n: break i = n//(n//(i + 1)) return result # Driver code N= 4print(totalSum(N)) N= 12print(totalSum(N)) |
15 127
Time complexity: O(log N)
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