bool IsPrime( int arr[], int size)
{
bool prime = true;
for( int i = 0; i < size; i++ )
{
for(int j = 2; j < arr[i]-1; j++)
{
if((arr[i] % j ) == 0 )
prime = false;
break;
}
if (prime)
return false;
}
}
void PrintPrime( int arr[], int size)
{
for ( int i = 0 ; i < size; i++ )
if ( IsPrime (arr,size) == false )
{
cout<<arr[i];
}
}
There are a few ways to calculate primes.
1) Write a recursive function that tries to devide a number by every number less that it.
2) (2^n)-1
2 to the power of n minus 1 is always a prime number, but you will skip some.
Best Regards,
Bonkey
Try again there, tiger. 2^4 = 16 - 1 = 15. There is no algebraic formula to compute the primes.2 to the power of n minus 1 is always a prime number, but you will skip some.
If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein
how about...
a_0 = 2
a_n = a_(n-1)*a_(n-2)*...*a_0 + 1
ex
a_1 = 2+1 = 3
a_2 = 3*2+1 = 7
a_3 = 7*3*2+1 = 43
a_4 = 43*7*3*2+1 = 1807
.
.
.
1807 = 13 * 139. No, really, there isn't one.
<snip>
a_4 = 43*7*3*2+1 = 1807
If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein
Not that it matters, but your equation isn't correct. You say that 2^4 = 15, which isn't trueTry again there, tiger. 2^4 = 16 - 1 = 15..
Anyway, there is no algebraic function to calculate the n:th prime number, but I'm pretty sure there is a function that can calculate a number which is prime (though it will skip a lot of prime numbers when doing so).
I'll reply if I find it.
MagosX.com
Give a man a fish and you feed him for a day.
Teach a man to fish and you feed him for a lifetime.
Aye...I should have split it up into two equations, but I was lazy.Originally posted by Magos
Not that it matters, but your equation isn't correct. You say that 2^4 = 15, which isn't true
I really think your search will be fruitless. People have been trying for centuries to find some pattern to the prime numbers. I think the fact that we're still trying says something about human intelligence.
If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein
thanks for you help
I am use the program I originally wrote..
Whats wrong with the code below..its not compiling
I the numbers aren't being passed correctly in the arrayCode:bool IsPrime( int arr *, int size) { { bool prime = true; for( int i = 0; i < size; i++ ) { for(int j = 2; j < arr[i]-1; j++) { if((arr[i] % j ) == 0 ) prime = false; break; } if (prime) return false; } } void PrintPrime( int arr *, int size) { { for ( int i = 0 ; i < size; i++ ) if ( IsPrime (arr,size) == false ) { cout<<arr[i]; } } }
so when primes print..I see all the numbers..instead of just the primes
[code][/code]tagged by Salem
Your IsPrime() function should look something like this.
And if you want to speed it up, the for-loop need only run to num/2 or even sqrt(num).Code:bool IsPrime(int num) { for (int j=2; j<=num-1; j++) { if (num%j == 0) return false; } return true; }
Thank You For your help
