C Program for Find sum of odd factors of a number
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 = 24
Input : 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:
C
#include <math.h>
#include <stdio.h>
int sumofoddFactors( int n)
{
int res = 1;
while (n % 2 == 0)
n = n / 2;
for ( int i = 3; i * i <= n; i++) {
int count = 0;
int curr_sum = 1;
int curr_term = 1;
while (n % i == 0) {
count++;
n = n / i;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2)
res *= (1 + n);
return res;
}
int main()
{
int n = 30;
printf ( "%d\n" , sumofoddFactors(n));
return 0;
}
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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!
Last Updated :
25 Oct, 2023
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