因此,我只是试图创建一个循环来运行“ 3n + 1”公式,当我输入一个负数时,我陷入了一个无限循环,余数为0和-1。

那是正确的还是我的代码缺少某些内容?

这是我的代码:

    Scanner scan = new Scanner(System.in);
    number = 0;
    method = 0;
int counter= 0;

if(scan.hasNextInt()){
     number = scan.nextInt();
  int original = number;
 while(number!=1){
      method = number%2;
     if(method==0){
    number = number/2;
 }else number = number*3+1;
 counter +=1;
  System.out.println(number);
  System.out.println("the remainder was "+method);
 }


 System.out.println("The original number was "+original);
 System.out.println("it took " + counter+ " times to reach 1.");



}else System.out.println("please enter a number");

最佳答案

该猜想仅适用于自然数(即正整数1、2、3,...)。如果要将其扩展为0和负数,则必须使用其他公式。在https://en.wikipedia.org/wiki/Collatz_conjecture上查看“扩展到更大的域”。

10-07 19:23
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