找到之前的斐波那契数
原文:https://www . geesforgeks . org/find-the-previous-Fibonacci-number/
给定一个斐波那契数 N ，任务是找到前一个斐波那契数。 例:
输入: N = 8 输出: 5 5 是 8 之前的前一个斐波那契数。 输入: N = 5 输出: 3
趋近:斐波那契数列中相邻两个数的比值迅速趋近 ((1 + sqrt(5)) / 2) 。所以如果 N 除以 ((1 + sqrt(5)) / 2) 再取整，结果数就是之前的斐波那契数。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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
java 描述语言
<script>
// 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.
</script>
Output:
5
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