找到下一个斐波那契数
原文:https://www . geesforgeks . org/find-the-next-Fibonacci-number/
给定一个斐波那契数 N ，任务是找到下一个斐波那契数。 例:
输入: N = 5 输出: 8 8 是 5 之后的下一个斐波那契数输入: N = 3 输出: 5
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) 。所以如果将 N 乘以 ((1 + sqrt(5)) / 2) ，取整，得到的数就是下一个斐波那契数。 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 next
// fibonacci number
int nextFibonacci(int n)
{
double a = n * (1 + sqrt(5)) / 2.0;
return round(a);
}
// Driver code
int main()
{
int n = 5;
cout << nextFibonacci(n);
}
// This code is contributed by mohit kumar 29
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the next
// fibonacci number
static long nextFibonacci(int n)
{
double a = n * (1 + Math.sqrt(5)) / 2.0;
return Math.round(a);
}
// Driver code
public static void main (String[] args)
{
int n = 5;
System.out.println(nextFibonacci(n));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
from math import *
# Function to return the next
# fibonacci number
def nextFibonacci(n):
a = n*(1 + sqrt(5))/2.0
return round(a)
# Driver code
n = 5
print(nextFibonacci(n))
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the next
// fibonacci number
static long nextFibonacci(int n)
{
double a = n * (1 + Math.Sqrt(5)) / 2.0;
return (long)Math.Round(a);
}
// Driver code
public static void Main(String[] args)
{
int n = 5;
Console.WriteLine(nextFibonacci(n));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the next
// fibonacci number
function nextFibonacci(n)
{
let a = n * (1 + Math.sqrt(5)) / 2.0;
return Math.round(a);
}
// Driver code
let n = 5;
document.write(nextFibonacci(n));
// This code is contributed by Mayank Tyagi
</script>
Output:
8
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