Prime Numbers Optimization in C [closed]

问题

I try to print the prime numbers; 2 to 1 million. But nothing printed on the console. Could you check my code? And how can I be this code more optimized?

Here's my code:

#include <stdio.h>
#include <math.h>

main()
{
   int num, sr, num2;

   for (num = 2; num <= 1000000; num++) {
      sr = (int) sqrt(num);
      for (num2 = 2; sr % num2 != 0; num2++) {
         if (sr == num2) {
            printf("%d\n", sr);
         }
      }
   }

}

回答1:

    #include <stdio.h>
    #include <math.h>

    int main(){
    int num, sr, num2;
    int isPrime = 1; // this is a check parameter; isPrime = 0 if number is not prime.
    for(num=2; num<=100; num++){
        sr = (int) sqrt(num);
        for(num2=2; num2 <= sr; num2++){
            //num2 <== sr to stop the innner loop
            if(num%num2 == 0){
                //printf("=========== %d|%d\n", num,num2); // uncomment this to see if a number is divisable
                isPrime = 0; // this number is not prime, cos num can be divided by num2
                break;
            }
        }
        if(isPrime){
            printf("Prime number is %d\n", num);
            isPrime = 1; // reset the check parameter
        }else{
            isPrime = 1; // reset the check parameter
        }
    }
         return 0;
    }

This code works. Since it works, i'll let you play with it and optimize it. If you can't let us know. We'll try to help you.

I like how you used sqrt to optimize the code.

回答2:

Did it compile?

line 4: main() should be int main()?

another thing: sr = 1. 1 modulo any number is 1.

and finally. sr will never be equal to num2, because sr is 1 and num2 is 2 or greater so it will never print anything.

this will get you into an infinite loop that does nothing

回答3:

If you wish to optimize it, you should use something like the sieve of eratosthenes.It is easy to operate on your data range. You can read more about it here : http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

回答4:

One optimization that you can use is the fact that all primes above 3 are of the form 6n+1 or 6n-1 and the fact that if a number is divisible by a prime, it is not a prime. Here is some code that uses that fact:

#include <stdio.h>
#include <math.h>

int is_prime(long num)
{
    int k = 1, a = 0, b = 0;
    long sr;
    switch(num)
        {
        case 1: return 0;
        case 2: return 1;
        case 3: return 1;
        case 4: return 0;
        case 5: return 1;
        case 6: return 0;
        case 7: return 1;
    }
    if (num % 2 == 0) return 0;
    if (num % 3 == 0) return 0;
    sr = (int) sqrt(num);
    while (b < sr) {
        a = (6 * k) - 1;
        b = (6 * k) + 1;
        if (num % a == 0)
            return 0;
        if (num % b == 0)
            return 0;
        k += 1;
    }
    return 1;
}

void main(void)
{
    int j;

    for (j = 0; j<100; j++){
        if (is_prime(j))
            printf("%d is a prime\n", j);
    }
}

This function returns 1 if num is a prime.

来源：https://stackoverflow.com/questions/15464677/prime-numbers-optimization-in-c

标签

primes