# C Program for efficiently print all prime factors of a given number

Given a number n, write an efficient function to print all prime factors of n. For example, if the input number is 12, then output should be “2 2 3”. And if the input number is 315, then output should be “3 3 5 7”.

Following are the steps to find all prime factors. **1)** While n is divisible by 2, print 2 and divide n by 2. **2)** After step 1, n must be odd. Now start a loop from i = 3 to square root of n. While i divides n, print i and divide n by i, increment i by 2 and continue. **3)** If n is a prime number and is greater than 2, then n will not become 1 by above two steps. So print n if it is greater than 2.

## C

`// C Program to print all prime factors` `#include <math.h>` `#include <stdio.h>` `// A function to print all prime factors of a given number n` `void` `primeFactors(` `int` `n)` `{` ` ` `// Print the number of 2s that divide n` ` ` `while` `(n % 2 == 0) {` ` ` `printf` `(` `"%d "` `, 2);` ` ` `n = n / 2;` ` ` `}` ` ` `// n must be odd at this point. So we can skip` ` ` `// one element (Note i = i +2)` ` ` `for` `(` `int` `i = 3; i <= ` `sqrt` `(n); i = i + 2) {` ` ` `// While i divides n, print i and divide n` ` ` `while` `(n % i == 0) {` ` ` `printf` `(` `"%d "` `, i);` ` ` `n = n / i;` ` ` `}` ` ` `}` ` ` `// This condition is to handle the case when n` ` ` `// is a prime number greater than 2` ` ` `if` `(n > 2)` ` ` `printf` `(` `"%d "` `, n);` `}` `/* Driver program to test above function */` `int` `main()` `{` ` ` `int` `n = 315;` ` ` `primeFactors(n);` ` ` `return` `0;` `}` |

**Output:**

3 3 5 7

**Time Complexity: **O(n^{1/2})

**Auxiliary Space: **O(1)

**How does this work?**

The steps 1 and 2 take care of composite numbers and step 3 takes care of prime numbers. To prove that the complete algorithm works, we need to prove that steps 1 and 2 actually take care of composite numbers. This is clear that step 1 takes care of even numbers. And after step 1, all remaining prime factor must be odd (difference of two prime factors must be at least 2), this explains why i is incremented by 2.

Now the main part is, the loop runs till square root of n not till. To prove that this optimization works, let us consider the following property of composite numbers. *Every composite number has at least one prime factor less than or equal to square root of itself.*

This property can be proved using counter statement. Let a and b be two factors of n such that a*b = n. If both are greater than √n, then a.b > √n, * √n, which contradicts the expression “a * b = n”.

In step 2 of the above algorithm, we run a loop and do following in loop

a) Find the least prime factor i (must be less than √n, )

b) Remove all occurrences i from n by repeatedly dividing n by i.

c) Repeat steps a and b for divided n and i = i + 2. The steps a and b are repeated till n becomes either 1 or a prime number.

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