Question

完美数字是一个数字，等于所有正数除数的总和，不包括在内。

对于我的作业，我试图编写一个程序来查找10000以下的所有四个完美数字，但是当我运行它时我的代码不起作用而且我不确定为什么（它只运行一两秒，然后说＃34;建立成功＆＃34;没有打印任何东西）。我将其包括在内，以及一些解释我思考过程的评论。有人可以帮助我，并告诉我它有什么问题吗？

public class HomeworkTwo {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {

        //Variable declaration

        int n;
        int possibleFactor;
        int factorSum=0;


        /**For each n, the program looks through all positive integers less than n  
           and tries to find factors of n. Once found, it adds them
           together and checks if the sum equals n. Then it repeats till n=9999. **/

        for (n=2; n<10000; n++) {
            for (possibleFactor = 1; possibleFactor < n; possibleFactor++) {
                if (n % possibleFactor == 0) {
                    factorSum = possibleFactor + factorSum;
                }

                //Is the number perfect? Printing
                if (factorSum == n) {
                    System.out.println(""+ n +" is a perfect number.");
                }
            }
        }
    }
}

Answer 1

您在第一个factorSum循环之前将0初始化为for，但在尝试每个新0时，您不会将其重置为n。这些因素不断增加，并且永远不会与要检查的数量相等。在0 n循环的开头将其重置为for。

此外，您可能希望在内部for循环之后但在外部for循环结束之前移动测试并打印数字为完美数字，否则它可能会打印超过必要的。

Answer 2

您的程序存在一些问题：

循环完因素后，您需要将factorSum重置为0
您应该在添加所有因素后检查factorSum == n，而不是在循环内部。
您只需要查看n/2;例如10永远不会被7整除。

这是生成的程序（格式稍微好一些）：

public class HomeworkTwo {

  /**
   * @param args
   *          the command line arguments
   */
  public static void main(String[] args) {

    // Variable declaration

    int n;
    int possibleFactor;
    int factorSum = 0;

    /**
     * For each n, the program looks through all positive integers less than n
     * and tries to find factors of n. Once found, it adds them together and
     * checks if the sum equals n. Then it repeats till n=9999.
     **/

    for (n = 2; n < 10000; n++) {
      factorSum = 0;
      for (possibleFactor = 1; possibleFactor <= n / 2; possibleFactor++) {
        if (n % possibleFactor == 0) {
          factorSum = possibleFactor + factorSum;
        }
      }
      // Is the number perfect? Printing
      if (factorSum == n) {
        System.out.println("" + n + " is a perfect number.");
      }
    }
  }
}

Answer 3

我想你已经做到了，但无论如何，代码中的主要问题是“factorSum”变量。检查每个数字后，应再次将其设置为0。另外，我用printf代替println，但它是一样的：

public static void main(String[] args) {
    int number = 0;
    int factor = 0;
    int factorSum = 0;

    for(number = 2; number < 10000; number++) { //for each number we will check the possible factors.
        factorSum = 0;

        for(factor = 1; factor < number; factor++)
            if((number % factor) == 0) { //if it is a factor, we add his value to factorSum.
                factorSum = factorSum + factor;
            }

        if(factorSum == number) {
            System.out.printf("The number: %d is a perfect number.\n", number);
        }
    }
}

Answer 4

你应该保持这样

for (n=2; n<10000; n++) {
    for (possibleFactor = 1; possibleFactor < n; possibleFactor++) {
        if (n % possibleFactor == 0) {
            factorSum = possibleFactor + factorSum;
        }
    }

    //Is the number perfect? Printing
    if (factorSum == n) {
        System.out.println(""+ n +" is a perfect number.");
    }
    factorSum = 0;
}

查找并打印10000以下的完美数字（Liang，Java简介，练习5.33）

4 个答案: