Algoritm numar prim

[b]int[/b] prim (int x)
{
[b]int[/b] i, y[b]=[/b]1; // presupui ca numarul e prim
for (i=2;i<=x/2;i++)
 if (x%i[b] ==[/b]0) { y=0; break;} // am gasit deja un divizor, nu are sens sa continuam
return y; // intoarcem 1 sau 0
}

Edited by msmihai, 28 October 2009 - 16:20.

Edited by SoR!n, 28 October 2009 - 16:23.

int prim(int x)
{
	  for(int i = 2; i <= sqrt(x); i++)
			if(i % 2 > 0 && x % i == 0)
				  return 1;
	  return 0;
}

Edited by Ph@ntom, 29 October 2009 - 15:33.

 Ph@ntom, on 29th October 2009, 15:32, said:

int prim(int x)
{
	  for(int i = 2; i <= sqrt(x); i++)
			if(i % 2 > 0 && x % i == 0)
				  return 1;
	  return 0;
}

int prim(int x)
{
	for(int i = 2; i <= int(sqrt(x)); i++)
		if(x % i == 0)
			return false;
	return true;
}

Edited by Ph@ntom, 29 October 2009 - 21:08.

int prim(int x)
{int i,[b]p=1[/b];
if (x==0 || x==1) p=0; \\asta ma rog nu prea conteaza
	for(int i = 2; i <= int(sqrt(x)) [b]&&p[/b]; i++)
		if(x % i == 0)
		   p=0;
	return p;
}

int Prim(long n)
{  long i;
	if(n==1||n==0) return 0;
	if(n>2&&n%2==0) return 0;
	for(i=3;i*i<n;i=i+2)
		 if(n%i==0)
			  return 0;
	return 1;
}

 NumeDeCod, on 29th October 2009, 16:54, said: