Dado un entero positivo N , la tarea es encontrar el número de Fibonacci más cercano al entero N dado . Si hay dos números de Fibonacci que tienen la misma diferencia de N , imprima el valor más pequeño.
Ejemplos:
Entrada: N = 20
Salida: 21
Explicación: El número de Fibonacci más cercano a 20 es 21.Entrada: N = 17
Salida: 13
Enfoque: siga los pasos a continuación para resolver el problema:
- Si N es igual a 0 , imprima 0 como resultado.
- Inicialice una variable, digamos ans , para almacenar el número de Fibonacci más cercano a N .
- Inicialice dos variables, digamos First como 0 y Second como 1 , para almacenar el primer y segundo término de la serie de Fibonacci .
- Almacene la suma de First y Second en una variable, digamos Third .
- Iterar hasta que el valor de Third sea como máximo N y realizar los siguientes pasos:
- Actualice el valor de Primero a Segundo y Segundo a Tercero .
- Almacene la suma de Primero y Segundo en la variable Tercero .
- Si la diferencia absoluta de Second y N es como máximo el valor de Third y N , actualice el valor de ans como Second .
- De lo contrario, actualice el valor de ans como Third .
- Después de completar los pasos anteriores, imprima el valor de ans como resultado.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the Fibonacci
// number which is nearest to N
void nearestFibonacci(int num)
{
// Base Case
if (num == 0) {
cout << 0;
return;
}
// Initialize the first & second
// terms of the Fibonacci series
int first = 0, second = 1;
// Store the third term
int third = first + second;
// Iterate until the third term
// is less than or equal to num
while (third <= num) {
// Update the first
first = second;
// Update the second
second = third;
// Update the third
third = first + second;
}
// Store the Fibonacci number
// having smaller difference with N
int ans = (abs(third - num)
>= abs(second - num))
? second
: third;
// Print the result
cout << ans;
}
// Driver Code
int main()
{
int N = 17;
nearestFibonacci(N);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to find the Fibonacci
// number which is nearest to N
static void nearestFibonacci(int num)
{
// Base Case
if (num == 0)
{
System.out.print(0);
return;
}
// Initialize the first & second
// terms of the Fibonacci series
int first = 0, second = 1;
// Store the third term
int third = first + second;
// Iterate until the third term
// is less than or equal to num
while (third <= num)
{
// Update the first
first = second;
// Update the second
second = third;
// Update the third
third = first + second;
}
// Store the Fibonacci number
// having smaller difference with N
int ans = (Math.abs(third - num) >=
Math.abs(second - num)) ?
second : third;
// Print the result
System.out.print(ans);
}
// Driver Code
public static void main (String[] args)
{
int N = 17;
nearestFibonacci(N);
}
}
// This code is contributed by AnkThon
Python3
# Python3 program for the above approach # Function to find the Fibonacci # number which is nearest to N def nearestFibonacci(num): # Base Case if (num == 0): print(0) return # Initialize the first & second # terms of the Fibonacci series first = 0 second = 1 # Store the third term third = first + second # Iterate until the third term # is less than or equal to num while (third <= num): # Update the first first = second # Update the second second = third # Update the third third = first + second # Store the Fibonacci number # having smaller difference with N if (abs(third - num) >= abs(second - num)): ans = second else: ans = third # Print the result print(ans) # Driver Code if __name__ == '__main__': N = 17 nearestFibonacci(N) # This code is contributed by SURENDRA_GANGWAR
C#
// C# program for the above approach
using System;
class GFG{
// Function to find the Fibonacci
// number which is nearest to N
static void nearestFibonacci(int num)
{
// Base Case
if (num == 0)
{
Console.Write(0);
return;
}
// Initialize the first & second
// terms of the Fibonacci series
int first = 0, second = 1;
// Store the third term
int third = first + second;
// Iterate until the third term
// is less than or equal to num
while (third <= num)
{
// Update the first
first = second;
// Update the second
second = third;
// Update the third
third = first + second;
}
// Store the Fibonacci number
// having smaller difference with N
int ans = (Math.Abs(third - num) >=
Math.Abs(second - num)) ?
second : third;
// Print the result
Console.Write(ans);
}
// Driver Code
public static void Main(string[] args)
{
int N = 17;
nearestFibonacci(N);
}
}
// This code is contributed by sanjoy_62
Javascript
<script>
// Javascript program for the above approach
// Function to find the Fibonacci
// number which is nearest to N
function nearestFibonacci(num)
{
// Base Case
if (num == 0) {
document.write(0);
return;
}
// Initialize the first & second
// terms of the Fibonacci series
let first = 0, second = 1;
// Store the third term
let third = first + second;
// Iterate until the third term
// is less than or equal to num
while (third <= num) {
// Update the first
first = second;
// Update the second
second = third;
// Update the third
third = first + second;
}
// Store the Fibonacci number
// having smaller difference with N
let ans = (Math.abs(third - num)
>= Math.abs(second - num))
? second
: third;
// Print the result
document.write(ans);
}
// Driver Code
let N = 17;
nearestFibonacci(N);
// This code is contributed by subhammahato348.
</script>
Producción:
13
Complejidad temporal: O(log N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por rohitsingh07052 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA