Java程序如何检查给定的数字是否是斐波那契数？

给定一个数字“n”，如何检查 n 是否为斐波那契数。前几个斐波那契数是 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ..

例子 ：

Input : 8
Output : Yes

Input : 34
Output : Yes

Input : 41
Output : No

以下是关于斐波那契数的有趣属性，它也可用于检查给定数是否为斐波那契数。
一个数字是斐波那契当且仅当 (5*n ² + 4) 或 (5*n ² – 4) 之一或两者是完美正方形（来源：Wiki）。

// Java program to check if x is a perfect square
  
class GFG {
    // A utility method that returns true if x is perfect square
    static boolean isPerfectSquare(int x)
    {
        int s = (int)Math.sqrt(x);
        return (s * s == x);
    }
  
    // Returns true if n is a Fibonacci Number, else false
    static boolean isFibonacci(int n)
    {
        // n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or both
        // is a perfect square
        return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);
    }
  
    // Driver method
    public static void main(String[] args)
    {
        for (int i = 1; i <= 10; i++)
            System.out.println(isFibonacci(i) ? i + " is a Fibonacci Number"
                                              : i + " is a not Fibonacci Number");
    }
}
// This code is contributed by Nikita Tiwari

输出：

1 is a Fibonacci Number
2 is a Fibonacci Number
3 is a Fibonacci Number
4 is a not Fibonacci Number
5 is a Fibonacci Number
6 is a not Fibonacci Number
7 is a not Fibonacci Number
8 is a Fibonacci Number
9 is a not Fibonacci Number
10 is a not Fibonacci Number

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