problem 10 done, and problem 7 rewritten (new is_prime algorithm)

p7.c

@@ -5,50 +5,25 @@
  * Projecteuler - Problem 7
  * ------------------------
  * 2009-12-28 Henrik Hautakoski
- * 
- * fast and elegant solution using The fundamental theorem of arithmetic wich state: 
- * any natural number greater than 1 can be prime factorized in only one unique way 
- * unless it is prime itself.
  */
 
 #include <stdio.h>
-#include <malloc.h>
-
-int isprime(int *pl, int l, int p) {
-    
-    int i;	
-	
-	for(i=0; i < l; i++) {
-
-		if((p % pl[i]) == 0)
-			return 0;
-	}
-	
-	return 1;
-}
+#include "lib/prime.h"
 
 int main() {
 	
-    int i, len = 3, *prime = malloc(sizeof(int)*10001);
+    uint32_t i, p, n = 1;
 	
-    if (prime == NULL)
-        return 1;
-    
-    prime[0] = 2;
-    prime[1] = 3;
-    prime[2] = 5;
-    
-	for(i=6; len < 10001; i++) {
-		
-		if(!isprime(prime, len, i))
-			continue;
-		
-		prime[len++] = i;
-	}
+    for(i=3; n < 10001; i+=2) {
 
-	printf("%i\n", prime[len-1]);
-    
-    free(prime);
+        if (!is_prime(i))
+            continue;
+
+        p = i;
+        n++;
+    }
+
+	printf("%i\n", p);
     
 	return 0;
 }