Printing of primary numbers from 2 to 1000

Question

I am writing code that writes all primes from 2 to 1000 in a file called primes.txt. For some reason, I cannot find the correct way to solve this problem.

import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;

public class Problem6 {

    /**
     * @param args
     * @throws FileNotFoundException 
     */
    public static void main(String[] args) throws FileNotFoundException {
        PrintWriter prw = new PrintWriter("primes.txt");
        for (int i = 2; i <= 1000; i++){
            if (checkIfPrime(i) == true){
                System.out.println(i);
                prw.println(i);
            }
        }
    }

    public static boolean checkIfPrime (int num){
        boolean isPrime = true;  
        for (int i = 2; i <= 1000; i++){
            if ( num % i == 0 ){
                isPrime = false;
            }
        }

        return isPrime;
    }
}

I just don’t know what to do ... Please help. Thanks!

Answer 1

What happens when you pass your first number,, 2on checkIfPrime? It takes 2 to 2, which is 0, falsely claiming to be 2not simple.

, num. i , i num. (, , i num).

for (int i = 2; i < num; i++){

for (int i = 2; i <= Math.sqrt(num); i++){

, , ( , 1000). - .

Answer 2

, . , .

    static List<Integer> primes = new ArrayList<Integer>();

public static void main(String[] args) {
    for (int i = 2; i < 10000; i++) {
        if(checkPrime(i)){
            primes.add(i);
        }
    }
    System.out.println(primes);
}

private static boolean checkPrime(int n) {
    for (Integer i : primes) {
        if(i*i > n ){
            break;
        }else if(n%i==0 )
            return false;
     }
    return true;
}

Answer 3

for checkIfPrime(int num)

for (int i = 2; i < num; i++) {

---

BTW if (checkIfPrime(i) == true){ if (checkIfPrime(i)){

Answer 4

num , , num. ?: -)

Answer 5

" " Eratoshenes 2-3-5-7 wheel 1000. C-like ,

primes_1000()
{
   // the 2-3-5-7 wheel
   int wh[48] = {10,2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,
                  2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2};
   // core primes' multiples, each with its pointer into the wheel
   int m[7][4] = { {1,11,11,11*11}, {2,13,13,13*13}, {3,17,17,17*17},
                   {4,19,19,19*19}, {5,23,23,23*23}, {6,29,29,29*29},
                   {7,31,31,31*31} };    // 23*23 = 529 
   int i=1, p=11, k=0;
   print(2); print(3); print(5); print(7);
   p = 11;             // first number on the wheel - the first candidate
   do {
      // the smallest duplicate multiple is 121*13, ==> no dups below 1000!
      for( k=0; k < 7; ++k) {
         if ( p == m[k][3] ) {             // p is a multiple of m[k][1] prime:
            m[k][2] += wh[ m[k][0]++ ];    //   next number on the wheel
            m[k][3]  = m[k][1] * m[k][2];  //   next multiple of m[k][1] 
            m[k][0] %= 48;                 //   index into the wheel
            break;
         }
      }
      if (k == 7) {    // multiple of no prime below 32 -
          print(p);    //   - a prime below 1000!   (32^2 = 1024)
      }
      p += wh[i++];    // next number on the candidates wheel
      i %= 48;         // wrap around to simulate circular list
   } while ( p < 1000 );
}

500 4 , {11,13,17,19} , 2,3,5,7.

(. 1 100).

m - (multiplesOf(p) = map( multiplyBy(p), rollWheelFrom(p) ), ). , min- .

m/p >

primes, , .