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Find Sum of Odd Factors of a Number in C++
Given with a positive integer and the task is to generate the odd factors of a number and finding out the sum of given odd factors.
Example
Input-: number = 20 Output-: sum of odd factors is: 6 Input-: number = 18 Output-: sum of odd factors is: 13
So, result = 1 + 5 = 6
Approach used in the below program is as follows −
- Input the number for calculating the sum of odd factors of that number
- Ignore the digit 0 and 2 because both are even digits and store the digit 1 because it is an odd digit
- Start the loop from 3 till square root of a number
- Traverse till number % i is returning 0 and keep dividing the number with the value of i
- In the loop keep setting the value of temporary variable to temp = temp * i
- Set the total to total + temp
- Return the value of final res variable and print the result
Algorithm
START Step 1-> Declare function to calculate sum of odd factors int sum(int num) declare int res = 1 Loop While(num % 2 == 0) set num = num / 2 End Loop For int i = 3 and i <= sqrt(num) and i++ declare int count = 0 and total = 1 declare int temp = 1 Loop while (num % i == 0) count++ set num = num / i set temp *= i set total += temp End set res = res*total End IF (num >= 2) set res *= (1 + num) End return res Step 2-> In main() Declare int num = 20 call sum(num) STOP
Example
#include <bits/stdc++.h> using namespace std; //calculate sum of odd factors int sum(int num) { int res = 1; while (num % 2 == 0) num = num / 2; for (int i = 3; i <= sqrt(num); i++) { int count = 0, total = 1 ; int temp = 1; while (num % i == 0) { count++; num = num / i; temp *= i; total += temp; } res = res*total; } if (num >= 2) res *= (1 + num); return res; } int main() { int num = 20; cout<<"sum of odd factors is : "; cout <<sum(num); return 0; }
Output
sum of odd factors is : 6
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