打印前N个素数

Question

语句是:编写一个读取整数N并打印前N个素数的程序。

public static void main(String[] args) 
{
    Scanner scan = new Scanner(System.in);

    int N = scan.nextInt();
    int x = 2;

    for(int i = 0; i <= N; i++)
    {
        int count = 0;

        for(int j = 1; j <= x; j++)
            if(x%j == 0)
                count++;

        if(count == 2)
            System.out.print(x + " ");

        x++;
    }
}

当我运行这段代码时，它没有给出确切的N个数字。例如，对于N=1 &2，它打印前2个质数，对于N=3& 4，它打印前3个质数，对于N=5& 6，它打印前4个质数，依此类推。这段代码有什么问题？

Answer 1

我认为您的程序中有许多缺陷需要修复，因此我决定编写一个更简单、更优雅的程序。

Scanner scan = new Scanner(System.in);
int N = Integer.parseInt( scan.nextLine());
int count = 0;
int num = 2;
while(count != N) { // while count!= number of prime numbers entered keep searching..
    boolean prime = true;// to determine whether the number is prime or not
    for (int i = 2; i <= Math.sqrt(num); i++) { //efficiency matters
        if (num % i == 0) {
            prime = false; // if number divides any other number its not a prime so set prime to false and break the loop.
            break;
        }

    }
    if (prime) {
        count++;
        System.out.println(num);
    }
    num++; see if next number is prime or not.
}

Answer 2

// Java Program to generate first 'N' Prime Numbers
public class prime
{
    public static void main(String[]args)
    {
        int count = 0, max_count = 100, i;
        System.out.println("First "+max_count+" Prime Numbers:");

        for(int num=1; count<max_count; num++)
        {
            for(i=2; num%i != 0; i++);

            if(i == num)
            {
                System.out.print(" "+num);
                count++;
            }
        }
    }
}

Answer 3

使用动态编程：

任何不是素数的数总是可以被数列中它前面的至少一个素数整除。如果任何一个数不能除以它前面的所有质数，那么这个数也是质数

void printPrimeNumbers(int n) {
    ArrayList<Integer> primeNumbers = new ArrayList<>();
    primeNumbers.add(2);
    for (int i = 3; i < n; i+=2) { // skip over even numbers since they are not prime
        boolean isPrime = true;
        for (Integer prime : primeNumbers) { // check current prime numbers to see if it evenly divides into number
            if (i % prime == 0) { // when number is not prime
                isPrime = false;
                break; // optimization: stop checking when number is already not prime
            }
        }
        if (isPrime) {
            primeNumbers.add(i);
        }
    }
    System.out.println(primeNumbers);
}