我的我的Fibonacci数字程序遇到了我的Intro to Java类。它仅在我按升序键入数字时才有效。
目标:必须使用while循环,必须检测数字是否为斐波纳契数,并且必须检测序列中数字的顺序。如果它不是斐波纳契数,它必须这样说,并告诉它之间的数字。在用户退出
之前,程序还必须继续询问输入编号输出的示例:
Please input a number for analysis >> 2
2 is a fibonacci number whose order in the sequence is 4
Please input a number for analysis >> 53
53 is not a fibonacci number
However, it lies between the Fibonacci numbers 34 (order:10) and 55 (order:11)
我的问题:只有当我按升序输入输入数字时,程序才有效。所以例如,如果我输入5,它告诉我它是带有seq的fib数。但是当我输入2时,它表示它不是一个原始数字,而是位于Fibonacci数字3(顺序:5)和5(顺序:6)之间。当我输入55时,它告诉我它是一个原始数字。但是如果没有显示不正确的输入,我就不能输入任何低于55的数字。
我认为问题出在else if语句中,可能是(fibnext!= testnum)。或者算一下?任何人都可以指出我可能出现问题的方向吗?
import java.util.Scanner;
public class Fibonacci
{
public Fibonacci()
{
int fibhigh = 1;
int fiblow = 0;
int count = 2;
int fibnext = 0;
Scanner input = new Scanner(System.in);
boolean quit = false;
System.out.println("Welcome to Fibonacci Sequence Dectector");
while(!false)
{
System.out.print("Please input a number for analysis >> ");
int testnum = input.nextInt();
if(testnum < 0)
{
System.out.println("Please enter a positive number");
continue;
}
while(fibnext < testnum)
{
fibnext = fibhigh + fiblow;
fiblow = fibhigh;
fibhigh = fibnext;
count++;
}
if(fibnext == testnum)
{
System.out.println(testnum + " is a fibonacci number whose order in the sequence is " + count);
}
else if(fibnext != testnum) System.out.println(testnum + " is not a fibonacci number \nHowever, it lies between the Fibonacci numbers " + fiblow + " (order:" + (count - 1) + ") and " + fibhigh + " (order:" + count + ")");
}
}
}
答案 0 :(得分:1)
你永远不会清除纤维,纤维和纤维。这意味着您每次都会比较最高的数字。您应该在用户输入数据之前将值设置为原始数字。
答案 1 :(得分:1)
你永远不会重置变量
int fibhigh = 1, fiblow = 0, count = 2, fibnext = 0;
您应该移动此代码段
int fibhigh = 1;
int fiblow = 0;
int count = 2;
int fibnext = 0;
在循环开始时:
while (!false) { }