Test if a number is fibonacci

Question

I know how to make the list of the Fibonacci numbers, but i don\'t know how can i test if a given number belongs to the fibonacci list - one way that comes in mind is genera

Answer 1

Java solution can be done as below. But still it can be optimized

The following solution works for

1. 1≤T≤10 ^5
2. 1≤N≤10 ^10

T is no.of test cases, N is range of number

    import java.util.Scanner;
    import java.math.BigDecimal;
    import java.math.RoundingMode;

    public class FibonacciTester {
        private static BigDecimal zero = BigDecimal.valueOf(0);
        private static BigDecimal one = BigDecimal.valueOf(1);
        private static BigDecimal two = BigDecimal.valueOf(2);
        private static BigDecimal four = BigDecimal.valueOf(4);
        private static BigDecimal five = BigDecimal.valueOf(5);

        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            int n = sc.nextInt();
            BigDecimal[] inputs = new BigDecimal[n];
            for (int i = 0; i < n; i++) {
                inputs[i] = sc.nextBigDecimal();
            }

            for (int i = 0; i < inputs.length; i++) {
                if (isFibonacci(inputs[i]))
                    System.out.println("IsFibo");
                else
                    System.out.println("IsNotFibo");
            }


        }

        public static boolean isFibonacci(BigDecimal num) {
            if (num.compareTo(zero) <= 0) {
                return false;
            }

            BigDecimal base = num.multiply(num).multiply(five);
            BigDecimal possibility1 = base.add(four);
            BigDecimal possibility2 = base.subtract(four);


            return (isPerfectSquare(possibility1) || isPerfectSquare(possibility2));
        }

        public static boolean isPerfectSquare(BigDecimal num) {
            BigDecimal squareRoot = one;
            BigDecimal square = one;
            BigDecimal i = one;
            BigDecimal newSquareRoot;
            int comparison = -1;

            while (comparison != 0) {
                if (comparison < 0) {
                    i = i.multiply(two);
                    newSquareRoot = squareRoot.add(i).setScale(0, RoundingMode.HALF_UP);
                } else {
                    i = i.divide(two);
                    newSquareRoot = squareRoot.subtract(i).setScale(0, RoundingMode.HALF_UP);
                }

                if (newSquareRoot.compareTo(squareRoot) == 0) {
                    return false;
                }

                squareRoot = newSquareRoot;
                square = squareRoot.multiply(squareRoot);
                comparison = square.compareTo(num);
            }

            return true;
        }
    }

Answer 2

int isfib(int n /* number */, int &pos /* position */)
{
   if (n == 1)
   {
      pos=2;  // 1 1
      return 1;
   }
   else if (n == 2)
   {
      pos=3;  // 1 1 2
      return 1;
   }
   else
   {
      int m = n /2;
      int p, q, x, y;
      int t1=0, t2 =0;
      for (int i = m; i < n; i++)
      {
        p = i;
        q = n -p;    // p + q = n
        t1 = isfib(p, x);
        if (t1) t2 = isfib(q, y);
        if (t1 && t2 && x == y +1)
        {
           pos = x+1;
           return 1; //true
        }
      }
      pos = -1;
      return 0; //false
   }
}

How about this?