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C++ program for Find sum of odd factors of a number
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
Updated on: 09-Jul-2020
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