Question

我制作了一个打印1-100个素数的程序。请帮我在1到100个数字的范围内抛出复合数的异常。我是初学者，所以任何帮助将不胜感激。

  public static void main(String[] args) {

    System.out.println("Prime numbers from 1 - 100 are :");
    int i = 0;
    int x = 0;

    for (i = 1; i <= 100; i++) {

        int ctr = 0;
        for (x = i; x >= 1; x--) {
            if (i % x == 0) {
                ctr = ctr + 1;
            }
        }
        if (ctr == 2) {
            System.out.println(i);
        }

    }

}

Answer 1

我宁愿实现isPrime方法并将其命名为

public static boolean isPrime(int value) {
  if (value <= 1)
    return false;

  // There's only one even prime: that is two 
  if ((value % 2) == 0) 
    return (value == 2);

  int from = (int) (Math.sqrt(value) + 1);

  // You have to check possible divisors from 3 to sqrt(value)
  for (int i = 3; i <= from; i += 2)
    if ((value % i) == 0)
      return false;

  return true;
}

public static void main(String[] args) {
  ...

  for (int i = 1; i <= 100; ++i) {
    if (isPrime(i)) 
      System.out.println(i);
    else {
      // i is not prime. You can do nothing, throw an exception etc
      // throw new MyException("Not a prime"); 
    }
  }
}

Answer 2

您应该向else添加if (ctr == 2) {子句并在其中引发异常。看看the documentation如何抛出异常。

Answer 3

再加上一个条件。否则if（ctr！= 2）{ 抛出新的CompositeException（“发生复合异常”）;}

Answer 4

public void primeNumber（int n1，int n2）{

    String primeNo= "";
    System.out.print("Prime number between " + n1 +" and "+ n2 + "is/are - ");
    for(int i = n1;i <=n2; i++)
    {
        int count = 0;
        for(int j=1; j<=i; j++)
        {

           if(i%j==0)
           {
               count = count + 1;
           }

        }
        if(count == 2) {
            primeNo = primeNo + i + ", ";
        }

    }

是System.out.print（primeNo）;

程序打印1-100素数并在给定范围内抛出复合数的异常

4 个答案: