查找以前的斐波那契数

给定斐波那契数N ，任务是找到先前的斐波那契数。
例子：

Input: N = 8 
Output: 5 
5 is the previous fibonacci number before 8.
Input: N = 5 
Output: 3 
 

方法：斐波那契数列中两个相邻数字的比值迅速接近((1 + sqrt(5))/ 2) 。因此，如果将N除以(((1 + sqrt(5))/ 2))然后四舍五入，则结果数将是先前的斐波那契数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
 
using namespace std;
 
// Function to return the previous
// fibonacci number
int previousFibonacci(int n)
{
    double a = n / ((1 + sqrt(5)) / 2.0);
    return round(a);
}
 
// Driver code
int main()
{
    int n = 8;
    cout << (previousFibonacci(n));
}
 
// This code is contributed by Mohit Kumar

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
         
// Function to return the previous
// fibonacci number
static int previousFibonacci(int n)
{
    double a = n / ((1 + Math.sqrt(5)) / 2.0);
    return (int)Math.round(a);
}
 
// Driver code
public static void main (String[] args)
{
    int n = 8;
    System.out.println(previousFibonacci(n));
}
}
 
// This code is contributed by ajit.

Python3

# Python3 implementation of the approach
from math import *
 
# Function to return the previous
# fibonacci number
def previousFibonacci(n):
    a = n/((1 + sqrt(5))/2.0)
    return round(a)
 
# Driver code
n = 8
print(previousFibonacci(n))

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the previous
// fibonacci number
static int previousFibonacci(int n)
{
    double a = n / ((1 + Math.Sqrt(5)) / 2.0);
    return (int)Math.Round(a);
}
 
// Driver code
public static void Main()
{
    int n = 8;
    Console.Write(previousFibonacci(n));
}
}
 
// This code is contributed by Akanksha_Rai

Javascript

// Javascript implementation of the approach
 
// Function to return the previous
// fibonacci number
function previousFibonacci(n)
{
    var a = n / ((1 + Math.sqrt(5)) / 2);
    return Math.round(a);
}
 
// Driver code
var n = 8;
document.write(previousFibonacci(n));
 
// This code is contributed by rutvik_56.

输出：

5