How can I find the prime numbers? - c++

Question

How can I find the prime numbers in a one-dimensional array in C++ in a simple way ??
{
int list[5];
int i,sum = 0;
for (i = 0; i < 5; i++)
{
cout << "Enter The List [" << i << "]: "; cin >> list[i];
sum = sum + list[i];
}
cout << endl;
cout << "The Sum Is:" << sum << endl;
}

Answer 1

Emphasizing on the comment of #john:
Create a function (say bool is_prime(int n)).
Now check if the number n is a prime or not.
So, you need to check if each of the positive integers more than 1 before n divides n or not without leaving any remainder. There's a shorter workaround, which will greatly reduce the computational cost. Just checking till the square root of the number n will do. Hence the function sqrt() is used.
So now, our is_prime() function is pretty easy to build as you can see:
bool is_prime(int n)
{
int i,p=0;
for(i=2;i<=sqrt(n);i++)
{
if(n%i==0)
{
p=1;
break; //even if one integer divides the number, then it is composite.
}
}
if(p==1)
return false; //The number is a composite.
else
return true; //The number is a prime.
}
Now, you just need to pass every value of the array into this function, and your job will be done.
Also, this program can be made even better if you check for the special case of 1 which is neither composite nor prime. A suggestion is, check your array element if it is 1 or not. If not, then pass the value in the function, else just print that it is a 1.
NOTE: The sqrt() function is available in the cmath library in C++ so you need to include that in your program too.

Answer 2

You can use sieve of Eratosthenes. Simply how it works is it iterates (from 2) through an boolean array and if arr[i] is prime (is true, i is the given number), sets every multiplicity to false.
Start with an array filled with true
Numbers 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
is prime 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
first you have to set arr[0] and arr[1] to false, because these are not prime numbers
Numbers 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
is prime 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Now, you go to 2 and set every multiplication of it to false.
in this case 4, 6, 8, 10, 12, 14 16...
Numbers 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
is prime 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0
Then do it for 3
so 6, 9, 12, 15
Numbers 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
is prime 0 0 1 1 0 1 0 1 0 0 0 1 0 1 0 0 0
4 is not prime so you skip it
5 is prime so you do the same as for 2 and 3 (10 -> false, 15 -> false etc.)
after you use it, you can simply check if n is prime
if (arr[n] == true)
cout << n << " is prime";
else
cout << n << " is not prime";
you can find it easily on internet, for example here (there are some optimizations you can add, too)