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”。