查找下一个非斐波那契数

给定数字N ，任务是找到下一个非斐波那契数。
例子：

Input: N = 4 
Output: 6 
6 is the next non-fibonacci number after 4
Input: N = 6 
Output: 7 
 

方法：由于斐波那契数列为

0, 1, 1, 2, 3, 5, 8, 13, 21, 34…. 
 

可以观察到，不存在任何两个连续的斐波那契数。因此，为了找到下一个非斐波那契数，会出现以下情况：

1. 如果N <= 3 ，则下一个非斐波那契数将是4
2. 如果N> 3 ，那么我们将检查(N + 1)是否为斐波那契数。
   • 如果(N + 1)是斐波那契数字，则(N + 2)将是下一个非斐波那契数字。
   • 否则(N +1)将成为答案

下面是上述方法的实现：

C++

// C++ implementation of the approach
 
#include 
using namespace std;
 
// Function to check if a
// number is perfect square
bool isPerfectSquare(int x)
{
    int s = sqrt(x);
    return (s * s == x);
}
 
// Function to check if a
// number is Fibinacci Number
bool isFibonacci(int N)
{
    // N is Fibinacci if either
    // (5*N*N + 4), (5*N*N - 4) or both
    // is a perferct square
    return isPerfectSquare(5 * N * N + 4)
           || isPerfectSquare(5 * N * N - 4);
}
 
// Function to find
// the next Non-Fibinacci Number
int nextNonFibonacci(int N)
{
 
    // Case 1
    // If N<=3, then 4 will be
    // next Non-Fibinacci Number
    if (N <= 3)
        return 4;
 
    // Case 2
    // If N+1 is Fibinacci, then N+2
    // will be next Non-Fibinacci Number
    if (isFibonacci(N + 1))
        return N + 2;
 
    // If N+1 is Non-Fibinacci, then N+2
    // will be next Non-Fibinacci Number
    else
        return N + 1;
}
 
// Driver code
int main()
{
    int N = 3;
    cout << nextNonFibonacci(N)
         << endl;
 
    N = 5;
    cout << nextNonFibonacci(N)
         << endl;
 
    N = 7;
    cout << nextNonFibonacci(N)
         << endl;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG{
  
// Function to check if a
// number is perfect square
static boolean isPerfectSquare(int x)
{
    int s = (int) Math.sqrt(x);
    return (s * s == x);
}
  
// Function to check if a
// number is Fibinacci Number
static boolean isFibonacci(int N)
{
    // N is Fibinacci if either
    // (5*N*N + 4), (5*N*N - 4) or both
    // is a perferct square
    return isPerfectSquare(5 * N * N + 4)
           || isPerfectSquare(5 * N * N - 4);
}
  
// Function to find
// the next Non-Fibinacci Number
static int nextNonFibonacci(int N)
{
  
    // Case 1
    // If N<=3, then 4 will be
    // next Non-Fibinacci Number
    if (N <= 3)
        return 4;
  
    // Case 2
    // If N+1 is Fibinacci, then N+2
    // will be next Non-Fibinacci Number
    if (isFibonacci(N + 1))
        return N + 2;
  
    // If N+1 is Non-Fibinacci, then N+2
    // will be next Non-Fibinacci Number
    else
        return N + 1;
}
  
// Driver code
public static void main(String[] args)
{
    int N = 3;
    System.out.print(nextNonFibonacci(N)
         +"\n");
  
    N = 5;
    System.out.print(nextNonFibonacci(N)
         +"\n");
  
    N = 7;
    System.out.print(nextNonFibonacci(N)
         +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python 3

# Python 3 implementation of the approach
from math import sqrt
 
# Function to check if a
# number is perfect square
def isPerfectSquare(x):
    s = sqrt(x)
    return (s * s == x)
 
# Function to check if a
# number is Fibinacci Number
def isFibonacci(N):
 
    # N is Fibinacci if either
    # (5*N*N + 4), (5*N*N - 4) or both
    # is a perferct square
    return isPerfectSquare(5 * N * N + 4) or \
            isPerfectSquare(5 * N * N - 4)
 
# Function to find
# the next Non-Fibinacci Number
def nextNonFibonacci(N):
     
    # Case 1
    # If N<=3, then 4 will be
    # next Non-Fibinacci Number
    if (N <= 3):
        return 4
 
    # Case 2
    # If N+1 is Fibinacci, then N+2
    # will be next Non-Fibinacci Number
    if (isFibonacci(N + 1)):
        return N + 2
 
    # If N+1 is Non-Fibinacci, then N+2
    # will be next Non-Fibinacci Number
    else:
        return N
 
# Driver code
if __name__ == '__main__':
    N = 3
    print(nextNonFibonacci(N))
    N = 4
    print(nextNonFibonacci(N))
 
    N = 7
    print(nextNonFibonacci(N))
     
# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG{
   
// Function to check if a
// number is perfect square
static bool isPerfectSquare(int x)
{
    int s = (int) Math.Sqrt(x);
    return (s * s == x);
}
   
// Function to check if a
// number is Fibinacci Number
static bool isFibonacci(int N)
{
    // N is Fibinacci if either
    // (5*N*N + 4), (5*N*N - 4) or both
    // is a perferct square
    return isPerfectSquare(5 * N * N + 4)
           || isPerfectSquare(5 * N * N - 4);
}
   
// Function to find
// the next Non-Fibinacci Number
static int nextNonFibonacci(int N)
{
   
    // Case 1
    // If N<=3, then 4 will be
    // next Non-Fibinacci Number
    if (N <= 3)
        return 4;
   
    // Case 2
    // If N+1 is Fibinacci, then N+2
    // will be next Non-Fibinacci Number
    if (isFibonacci(N + 1))
        return N + 2;
   
    // If N+1 is Non-Fibinacci, then N+2
    // will be next Non-Fibinacci Number
    else
        return N + 1;
}
   
// Driver code
public static void Main(String[] args)
{
    int N = 3;
    Console.Write(nextNonFibonacci(N)
         +"\n");
   
    N = 5;
    Console.Write(nextNonFibonacci(N)
         +"\n");
   
    N = 7;
    Console.Write(nextNonFibonacci(N)
         +"\n");
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript implementation of the approach
 
// Function to check if a
// number is perfect square
function isPerfectSquare(x)
{
    var s = parseInt(Math.sqrt(x));
    return (s * s == x);
}
 
// Function to check if a
// number is Fibinacci Number
function isFibonacci(N)
{
    // N is Fibinacci if either
    // (5*N*N + 4), (5*N*N - 4) or both
    // is a perferct square
    return isPerfectSquare(5 * N * N + 4)
           || isPerfectSquare(5 * N * N - 4);
}
 
// Function to find
// the next Non-Fibinacci Number
function nextNonFibonacci(N)
{
 
    // Case 1
    // If N<=3, then 4 will be
    // next Non-Fibinacci Number
    if (N <= 3)
        return 4;
 
    // Case 2
    // If N+1 is Fibinacci, then N+2
    // will be next Non-Fibinacci Number
    if (isFibonacci(N + 1))
        return N + 2;
 
    // If N+1 is Non-Fibinacci, then N+2
    // will be next Non-Fibinacci Number
    else
        return N + 1;
}
 
// Driver code
var N = 3;
document.write(nextNonFibonacci(N)+"<br>");
N = 5;
document.write(nextNonFibonacci(N)+"<br>");
N = 7;
document.write(nextNonFibonacci(N)+"<br>");
 
// This code is contributed by rutvik_56.

输出：

4
6
9