Divided C and C++ files into seperate folders

1 files changed, 0 insertions, 49 deletions

diff --git a/C_C++/IsPrime.c b/C_C++/IsPrime.c
deleted file mode 100644
index c7c7630..0000000
--- a/C_C++/IsPrime.c
+++ /dev/null
@@ -1,49 +0,0 @@
-#include <stdio.h>
-#include <stdbool.h>
-#include <math.h>
-
-bool isPrime(unsigned long num, unsigned long *divisible) {
-	// The formula 6n +- 1, explained in the comment below, only finds primes above 3
-	if (num <= 1) return false;
-	if (num <= 3) return true;
-
-	if (num % 2 == 0) {
-		*divisible = 2;
-		return false;
-	}
-	if (num % 3 == 0) {
-		*divisible = 3;
-		return false;
-	}
-
-	// All prime numbers follow the formula 6n +- 1 (where n >= 1). All composite numbers are made up of prime numbers, so this way, we skip iterating through a lot of nums.
-	// The biggest number we need to check is the square root of num. Otherwise, we're gonna have situations like 5 * 20 and 20 * 5.
-	// Source (didn't look at example code): https://en.wikipedia.org/wiki/Primality_test#Simple_methods
-
-	// It must be n - 1 <= max, and not n + 1 <= max, because there are some cases, where n + 1 > sqrt(num) but n - 1 == sqrt(num)
-	// For example: 289, 323, 2209, 10201, 22201
-	for (unsigned long n = 6, max = sqrt(num); n - 1 <= max; n += 6) {
-		if (num % (n - 1) == 0) {
-			*divisible = n - 1;
-			return false;
-		}
-		if (num % (n + 1) == 0) {
-			*divisible = n + 1;
-			return false;
-		}
-	}
-	return true;
-}
-
-int main() {
-	char numberInput[20];
-	printf("Natural number, up to 2^64-1: ");
-	fgets(numberInput, 20, stdin);
-
-	unsigned long number = 0, divisible = -1;
-	sscanf(numberInput, "%lu", &number);
-	if (isPrime(number, &divisible))
-		printf("Your number is prime!");
-	else
-		printf("Your number is not prime! It is divisible by %ld!\n", divisible);
-}