Write a C program for a given number n, the task is to find the odd factor sum.
Examples:
Input : n = 30
Output : 24
Explanation: Odd dividers sum 1 + 3 + 5 + 15 = 24Input : 18
Output : 13
Explanation: Odd dividers sum 1 + 3 + 9 = 13
C Program for Find sum of odd factors of a number using Prime Factorization:
To find sum of odd factors, we simply need to ignore even factors and their powers. For example, consider n = 18. It can be written as 2132 and sum of all factors is (1)*(1 + 2)*(1 + 3 + 32). Sum of odd factors (1)*(1+3+32) = 13.
To remove all even factors, we repeatedly divide n while it is divisible by 2. After this step, we only get odd factors. Note that 2 is the only even prime.
Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1a1)*(p2a2)* … (pkak).
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak)
Below is the Implementation of the above approach:
#include <math.h> #include <stdio.h> // Function to calculate the sum of all factors of 'n' int sumofoddFactors( int n)
{ // Initialize the result to 1
int res = 1;
// Ignore even factors by removing all powers of 2
while (n % 2 == 0)
n = n / 2;
for ( int i = 3; i * i <= n; i++) {
// Initialize a count for the current prime factor
int count = 0;
// Initialize the current sum for the prime factor
int curr_sum = 1;
// Initialize the current term for the prime factor
int curr_term = 1;
while (n % i == 0) {
// Increment the count for this prime factor
count++;
// Remove the factor from 'n'
n = n / i;
// Update the current term
curr_term *= i;
// Add the current term to the current sum
curr_sum += curr_term;
}
// Multiply the result by the current sum
res *= curr_sum;
}
// This condition is to handle the case when 'n' is a
// prime number.
if (n >= 2)
res *= (1 + n);
// Return the final result
return res;
} int main()
{ // The input number 'n'
int n = 30;
// Calculate and print the sum of odd factors
printf ( "%d\n" , sumofoddFactors(n));
// Return 0 to indicate successful execution
return 0;
} |
24
Time Complexity: O(n1/2)
Auxiliary Space: O(1)
Please refer complete article on Find sum of odd factors of a number for more details!