Числа Фибоначчи — это числа в следующей целочисленной последовательности.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..
В математических терминах последовательность Fn чисел Фибоначчи определяется рекуррентным соотношением
Fn = Fn-1 + Fn-2с начальными значениями
F0 = 0 и F1 = 1.
Учитывая число n, выведите n-е число Фибоначчи.
Примеры:
Вход : n = 2 Выход : 1 Вход : n = 9 Выход : 34
Напишите функцию int fib(int n), которая возвращает Fn. Например, если n = 0, то функция fib() должна возвращать 0. Если n = 1, то он должен возвращать 1. Для n > 1 он должен возвращать F>n-1 + Fn-2
Для n = 9 Выход:34
Ниже приведены различные методы получения n-го числа Фибоначчи.
Метод 1 (Используйте рекурсию)
Простой метод, представляющий собой прямую рекурсивную реализацию математического рекуррентного соотношения, приведенного выше.
//Fibonacci Series using Recursion
#include<bits/stdc++.h>
using namespace std;
int fib(int n)
{
if (n <= 1)
return n;
return fib(n-1) + fib(n-2);
}
int main ()
{
int n = 9;
cout << fib(n);
getchar();
return 0;
}
// This code is contributed
// by Akanksha Rai
//Fibonacci Series using Recursion
#include<stdio.h>
int fib(int n)
{
if (n <= 1)
return n;
return fib(n-1) + fib(n-2);
}
int main ()
{
int n = 9;
printf("%d", fib(n));
getchar();
return 0;
}
//Fibonacci Series using Recursion
class fibonacci
{
static int fib(int n)
{
if (n <= 1)
return n;
return fib(n-1) + fib(n-2);
}
public static void main (String args[])
{
int n = 9;
System.out.println(fib(n));
}
}
/* This code is contributed by Rajat Mishra */
# Function for nth Fibonacci number
def fib(n,st,en):
if n-1==0:
return enPytho
if n>0:
temp_st=en
en=st+en
return fib(n-1,temp_st,en)
if __name__ == "__main__":
print(fib(9,0,1))
#This code is contributed by Bhawna Priya
// C# program for Fibonacci Series
// using Recursion
using System;
public class GFG
{
public static int Fib(int n)
{
if (n <= 1)
{
return n;
}
else
{
return Fib(n - 1) + Fib(n - 2);
}
}
// driver code
public static void Main(string[] args)
{
int n = 9;
Console.Write(Fib(n));
}
}
// This code is contributed by Sam007
<?php
// Fibonacci Series
// using Recursion
// function returns
// the Fibonacci number
function fib($n)
{
if ($n <= 1)
return $n;
return fib($n - 1) +
fib($n - 2);
}
// Driver Code
$n = 9;
echo fib($n);
// This code is contributed by aj_36
?>
<script>
//Fibonacci Series using Recursion
let n = 9;
// function returns the Fibonacci number
function fib(n) {
if (n <= 1)
return n;
return fib(n-1) + fib(n-2);
}
//function call
document.write(fib(n));
//This code is contributed by Surbhi Tyagi
</script>
Выход:
34Временная сложность:T(n) = T(n) , которая является линейной.
Если бы исходное дерево рекурсии должно было быть реализовано, то это было бы дерево, но теперь в n раз вызывается функция рекурсии
Исходное дерево для рекурсии
fib(5) / \ fib(4) fib(3) / \ / \ fib(3) fib(2) fib(2) fib(1) / \ / \ / \ fib(2) fib(1) fib(1) fib(0) fib(1) fib(0) / \ fib(1) fib(0)
Оптимизированное дерево для рекурсии для кода выше
fib(5)
fib(4)
fib(3)
fib(2)
fib(1)
Дополнительное Пространство: O(n), если мы учитываем размер стека вызовов функций, в противном случае O(1).
Метод 2 (Используйте Динамическое программирование)
Мы можем избежать повторной работы, проделанной в методе 1, сохранив числа Фибоначчи, вычисленные до сих пор.
// C++ program for Fibonacci Series
// using Dynamic Programming
#include<bits/stdc++.h>
using namespace std;
class GFG{
public:
int fib(int n)
{
// Declare an array to store
// Fibonacci numbers.
// 1 extra to handle
// case, n = 0
int f[n + 2];
int i;
// 0th and 1st number of the
// series are 0 and 1
f[0] = 0;
f[1] = 1;
for(i = 2; i <= n; i++)
{
//Add the previous 2 numbers
// in the series and store it
f[i] = f[i - 1] + f[i - 2];
}
return f[n];
}
};
// Driver code
int main ()
{
GFG g;
int n = 9;
cout << g.fib(n);
return 0;
}
// This code is contributed by SoumikMondal
//Fibonacci Series using Dynamic Programming
#include<stdio.h>
int fib(int n)
{
/* Declare an array to store Fibonacci numbers. */
int f[n+2]; // 1 extra to handle case, n = 0
int i;
/* 0th and 1st number of the series are 0 and 1*/
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++)
{
/* Add the previous 2 numbers in the series
and store it */
f[i] = f[i-1] + f[i-2];
}
return f[n];
}
int main ()
{
int n = 9;
printf("%d", fib(n));
getchar();
return 0;
}
// Fibonacci Series using Dynamic Programming
class fibonacci
{
static int fib(int n)
{
/* Declare an array to store Fibonacci numbers. */
int f[] = new int[n+2]; // 1 extra to handle case, n = 0
int i;
/* 0th and 1st number of the series are 0 and 1*/
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++)
{
/* Add the previous 2 numbers in the series
and store it */
f[i] = f[i-1] + f[i-2];
}
return f[n];
}
public static void main (String args[])
{
int n = 9;
System.out.println(fib(n));
}
}
/* This code is contributed by Rajat Mishra */
# Fibonacci Series using Dynamic Programming
def fibonacci(n):
# Taking 1st two fibonacci numbers as 0 and 1
f = [0, 1]
for i in range(2, n+1):
f.append(f[i-1] + f[i-2])
return f[n]
print(fibonacci(9))
// C# program for Fibonacci Series
// using Dynamic Programming
using System;
class fibonacci {
static int fib(int n)
{
// Declare an array to
// store Fibonacci numbers.
// 1 extra to handle
// case, n = 0
int []f = new int[n + 2];
int i;
/* 0th and 1st number of the
series are 0 and 1 */
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++)
{
/* Add the previous 2 numbers
in the series and store it */
f[i] = f[i - 1] + f[i - 2];
}
return f[n];
}
// Driver Code
public static void Main ()
{
int n = 9;
Console.WriteLine(fib(n));
}
}
// This code is contributed by anuj_67.
<?php
//Fibonacci Series using Dynamic
// Programming
function fib( $n)
{
/* Declare an array to store
Fibonacci numbers. */
// 1 extra to handle case,
// n = 0
$f = array();
$i;
/* 0th and 1st number of the
series are 0 and 1*/
$f[0] = 0;
$f[1] = 1;
for ($i = 2; $i <= $n; $i++)
{
/* Add the previous 2
numbers in the series
and store it */
$f[$i] = $f[$i-1] + $f[$i-2];
}
return $f[$n];
}
$n = 9;
echo fib($n);
// This code is contributed by
// anuj_67.
?>
<script>
// Fibonacci Series using Dynamic Programming
function fib(n)
{
/* Declare an array to store Fibonacci numbers. */
let f = new Array(n+2); // 1 extra to handle case, n = 0
let i;
/* 0th and 1st number of the series are 0 and 1*/
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++)
{
/* Add the previous 2 numbers in the series
and store it */
f[i] = f[i-1] + f[i-2];
}
return f[n];
}
let n=9;
document.write(fib(n));
// This code is contributed by avanitrachhadiya2155
</script>
Выход:
34Метод 3 (Метод 2, оптимизированный для пространства)
Мы можем оптимизировать пространство, используемое в методе 2, сохранив два предыдущих числа только потому, что это все, что нам нужно, чтобы последовательно получить следующее число Фибоначчи.
// Fibonacci Series using Space Optimized Method
#include<bits/stdc++.h>
using namespace std;
int fib(int n)
{
int a = 0, b = 1, c, i;
if( n == 0)
return a;
for(i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return b;
}
// Driver code
int main()
{
int n = 9;
cout << fib(n);
return 0;
}
// This code is contributed by Code_Mech
// Fibonacci Series using Space Optimized Method
#include<stdio.h>
int fib(int n)
{
int a = 0, b = 1, c, i;
if( n == 0)
return a;
for (i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return b;
}
int main ()
{
int n = 9;
printf("%d", fib(n));
getchar();
return 0;
}
// Java program for Fibonacci Series using Space
// Optimized Method
class fibonacci
{
static int fib(int n)
{
int a = 0, b = 1, c;
if (n == 0)
return a;
for (int i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return b;
}
public static void main (String args[])
{
int n = 9;
System.out.println(fib(n));
}
}
// This code is contributed by Mihir Joshi
# Function for nth fibonacci number - Space Optimisation
# Taking 1st two fibonacci numbers as 0 and 1
def fibonacci(n):
a = 0
b = 1
if n < 0:
print("Incorrect input")
elif n == 0:
return a
elif n == 1:
return b
else:
for i in range(2,n+1):
c = a + b
a = b
b = c
return b
# Driver Program
print(fibonacci(9))
#This code is contributed by Saket Modi
// C# program for Fibonacci Series
// using Space Optimized Method
using System;
namespace Fib
{
public class GFG
{
static int Fib(int n)
{
int a = 0, b = 1, c = 0;
// To return the first Fibonacci number
if (n == 0) return a;
for (int i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return b;
}
// Driver function
public static void Main(string[] args)
{
int n = 9;
Console.Write("{0} ", Fib(n));
}
}
}
// This code is contributed by Sam007.
<?php
// PHP program for Fibonacci Series
// using Space Optimized Method
function fib( $n)
{
$a = 0;
$b = 1;
$c;
$i;
if( $n == 0)
return $a;
for($i = 2; $i <= $n; $i++)
{
$c = $a + $b;
$a = $b;
$b = $c;
}
return $b;
}
// Driver Code
$n = 9;
echo fib($n);
// This code is contributed by anuj_67.
?>
<script>
// Javascript program for Fibonacci Series using Space Optimized Method
function fib(n)
{
let a = 0, b = 1, c, i;
if( n == 0)
return a;
for(i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return b;
}
// Driver code
let n = 9;
document.write(fib(n));
// This code is contributed by Mayank Tyagi
</script>
Выход:
34Временная сложность: O(n)
Дополнительное пространство: O(1)
Метод 4 (использование мощности матрицы {{1, 1}, {1, 0}})
это еще порядка O(N), которое полагается на то, что если мы N раз умножить матрицу м = {{1,1},{1,0}} для себя (другими словами рассчитать мощность(М, N)), тогда мы получим (П+1) — му числу Фибоначчи как элемент в строке и столбце (0, 0) в результирующей матрицы.
Представленность матрицы дает следующие закрытые выражение для чисел Фибоначчи :
#include<bits/stdc++.h>
using namespace std;
// Helper function that multiplies 2
// matrices F and M of size 2*2, and
// puts the multiplication result
// back to F[][]
void multiply(int F[2][2], int M[2][2]);
// Helper function that calculates F[][]
// raise to the power n and puts the
// result in F[][]
// Note that this function is designed
// only for fib() and won't work as
// general power function
void power(int F[2][2], int n);
int fib(int n)
{
int F[2][2] = { { 1, 1 }, { 1, 0 } };
if (n == 0)
return 0;
power(F, n - 1);
return F[0][0];
}
void multiply(int F[2][2], int M[2][2])
{
int x = F[0][0] * M[0][0] +
F[0][1] * M[1][0];
int y = F[0][0] * M[0][1] +
F[0][1] * M[1][1];
int z = F[1][0] * M[0][0] +
F[1][1] * M[1][0];
int w = F[1][0] * M[0][1] +
F[1][1] * M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
void power(int F[2][2], int n)
{
int i;
int M[2][2] = { { 1, 1 }, { 1, 0 } };
// n - 1 times multiply the
// matrix to {{1,0},{0,1}}
for(i = 2; i <= n; i++)
multiply(F, M);
}
// Driver code
int main()
{
int n = 9;
cout << " " << fib(n);
return 0;
}
// This code is contributed by shivanisinghss2110
#include <stdio.h>
/* Helper function that multiplies 2 matrices F and M of size 2*2, and
puts the multiplication result back to F[][] */
void multiply(int F[2][2], int M[2][2]);
/* Helper function that calculates F[][] raise to the power n and puts the
result in F[][]
Note that this function is designed only for fib() and won't work as general
power function */
void power(int F[2][2], int n);
int fib(int n)
{
int F[2][2] = {{1,1},{1,0}};
if (n == 0)
return 0;
power(F, n-1);
return F[0][0];
}
void multiply(int F[2][2], int M[2][2])
{
int x = F[0][0]*M[0][0] + F[0][1]*M[1][0];
int y = F[0][0]*M[0][1] + F[0][1]*M[1][1];
int z = F[1][0]*M[0][0] + F[1][1]*M[1][0];
int w = F[1][0]*M[0][1] + F[1][1]*M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
void power(int F[2][2], int n)
{
int i;
int M[2][2] = {{1,1},{1,0}};
// n - 1 times multiply the matrix to {{1,0},{0,1}}
for (i = 2; i <= n; i++)
multiply(F, M);
}
/* Driver program to test above function */
int main()
{
int n = 9;
printf("%d", fib(n));
getchar();
return 0;
}
class fibonacci
{
static int fib(int n)
{
int F[][] = new int[][]{{1,1},{1,0}};
if (n == 0)
return 0;
power(F, n-1);
return F[0][0];
}
/* Helper function that multiplies 2 matrices F and M of size 2*2, and
puts the multiplication result back to F[][] */
static void multiply(int F[][], int M[][])
{
int x = F[0][0]*M[0][0] + F[0][1]*M[1][0];
int y = F[0][0]*M[0][1] + F[0][1]*M[1][1];
int z = F[1][0]*M[0][0] + F[1][1]*M[1][0];
int w = F[1][0]*M[0][1] + F[1][1]*M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
/* Helper function that calculates F[][] raise to the power n and puts the
result in F[][]
Note that this function is designed only for fib() and won't work as general
power function */
static void power(int F[][], int n)
{
int i;
int M[][] = new int[][]{{1,1},{1,0}};
// n - 1 times multiply the matrix to {{1,0},{0,1}}
for (i = 2; i <= n; i++)
multiply(F, M);
}
/* Driver program to test above function */
public static void main (String args[])
{
int n = 9;
System.out.println(fib(n));
}
}
/* This code is contributed by Rajat Mishra */
# Helper function that multiplies
# 2 matrices F and M of size 2*2,
# and puts the multiplication
# result back to F[][]
# Helper function that calculates
# F[][] raise to the power n and
# puts the result in F[][]
# Note that this function is
# designed only for fib() and
# won't work as general
# power function
def fib(n):
F = [[1, 1],
[1, 0]]
if (n == 0):
return 0
power(F, n - 1)
return F[0][0]
def multiply(F, M):
x = (F[0][0] * M[0][0] +
F[0][1] * M[1][0])
y = (F[0][0] * M[0][1] +
F[0][1] * M[1][1])
z = (F[1][0] * M[0][0] +
F[1][1] * M[1][0])
w = (F[1][0] * M[0][1] +
F[1][1] * M[1][1])
F[0][0] = x
F[0][1] = y
F[1][0] = z
F[1][1] = w
def power(F, n):
M = [[1, 1],
[1, 0]]
# n - 1 times multiply the
# matrix to {{1,0},{0,1}}
for i in range(2, n + 1):
multiply(F, M)
# Driver Code
if __name__ == "__main__":
n = 9
print(fib(n))
# This code is contributed
# by ChitraNayal
using System;
class GFG {
static int fib(int n)
{
int [,]F = new int[,] {{1, 1},
{1, 0} };
if (n == 0)
return 0;
power(F, n-1);
return F[0,0];
}
/* Helper function that multiplies 2
matrices F and M of size 2*2, and puts
the multiplication result back to F[][] */
static void multiply(int [,]F, int [,]M)
{
int x = F[0,0]*M[0,0] + F[0,1]*M[1,0];
int y = F[0,0]*M[0,1] + F[0,1]*M[1,1];
int z = F[1,0]*M[0,0] + F[1,1]*M[1,0];
int w = F[1,0]*M[0,1] + F[1,1]*M[1,1];
F[0,0] = x;
F[0,1] = y;
F[1,0] = z;
F[1,1] = w;
}
/* Helper function that calculates F[][]
raise to the power n and puts the result
in F[][] Note that this function is designed
only for fib() and won't work as general
power function */
static void power(int [,]F, int n)
{
int i;
int [,]M = new int[,]{{1, 1},
{1, 0} };
// n - 1 times multiply the matrix to
// {{1,0},{0,1}}
for (i = 2; i <= n; i++)
multiply(F, M);
}
/* Driver program to test above function */
public static void Main ()
{
int n = 9;
Console.WriteLine(fib(n));
}
}
// This code is contributed by anuj_67.
<?php
function fib($n)
{
$F = array(array(1, 1),
array(1, 0));
if ($n == 0)
return 0;
power($F, $n - 1);
return $F[0][0];
}
function multiply(&$F, &$M)
{
$x = $F[0][0] * $M[0][0] +
$F[0][1] * $M[1][0];
$y = $F[0][0] * $M[0][1] +
$F[0][1] * $M[1][1];
$z = $F[1][0] * $M[0][0] +
$F[1][1] * $M[1][0];
$w = $F[1][0] * $M[0][1] +
$F[1][1] * $M[1][1];
$F[0][0] = $x;
$F[0][1] = $y;
$F[1][0] = $z;
$F[1][1] = $w;
}
function power(&$F, $n)
{
$M = array(array(1, 1),
array(1, 0));
// n - 1 times multiply the
// matrix to {{1,0},{0,1}}
for ($i = 2; $i <= $n; $i++)
multiply($F, $M);
}
// Driver Code
$n = 9;
echo fib($n);
// This code is contributed
// by ChitraNayal
?>
<script>
// Note that this function is designed
// only for fib() and won't work as
// general power function
function fib( n)
{
var F = [ [ 1, 1 ], [ 1, 0 ] ];
if (n == 0)
return 0;
power(F, n - 1);
return F[0][0];
}
// Helper function that multiplies 2
// matrices F and M of size 2*2, and
// puts the multiplication result
// back to F[][]
function multiply( F, M )
{
x = F[0][0] * M[0][0] +
F[0][1] * M[1][0];
y = F[0][0] * M[0][1] +
F[0][1] * M[1][1];
z = F[1][0] * M[0][0] +
F[1][1] * M[1][0];
w = F[1][0] * M[0][1] +
F[1][1] * M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
// Helper function that calculates F[][]
// raise to the power n and puts the
// result in F[][]
function power( F, n)
{
var i;
var M = [[ 1, 1 ], [ 1, 0 ]];
// n - 1 times multiply the
// matrix to {{1,0},{0,1}}
for(i = 2; i <= n; i++)
multiply(F, M);
}
// Driver code
var n = 9;
document.write (" " + fib(n));
//This code is contributed by sweetyty
</script>
Выход:
34Временная сложность: O(n)
Дополнительное пространство: O(1)
Метод 5 (Оптимизированный метод 4)
Метод 4 может быть оптимизирован для работы в O(логарифмической) временной сложности. Мы можем выполнить рекурсивное умножение, чтобы получить мощность(M, n) в предыдущем методе.
// Fibonacci Series using Optimized Method
#include <bits/stdc++.h>
using namespace std;
void multiply(int F[2][2], int M[2][2]);
void power(int F[2][2], int n);
// Function that returns nth Fibonacci number
int fib(int n)
{
int F[2][2] = {{1, 1}, {1, 0}};
if (n == 0)
return 0;
power(F, n - 1);
return F[0][0];
}
// Optimized version of power() in method 4
void power(int F[2][2], int n)
{
if(n == 0 || n == 1)
return;
int M[2][2] = {{1, 1}, {1, 0}};
power(F, n / 2);
multiply(F, F);
if (n % 2 != 0)
multiply(F, M);
}
void multiply(int F[2][2], int M[2][2])
{
int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
// Driver code
int main()
{
int n = 9;
cout << fib(9);
getchar();
return 0;
}
// This code is contributed by Nidhi_biet
#include <stdio.h>
void multiply(int F[2][2], int M[2][2]);
void power(int F[2][2], int n);
/* function that returns nth Fibonacci number */
int fib(int n)
{
int F[2][2] = {{1,1},{1,0}};
if (n == 0)
return 0;
power(F, n-1);
return F[0][0];
}
/* Optimized version of power() in method 4 */
void power(int F[2][2], int n)
{
if( n == 0 || n == 1)
return;
int M[2][2] = {{1,1},{1,0}};
power(F, n/2);
multiply(F, F);
if (n%2 != 0)
multiply(F, M);
}
void multiply(int F[2][2], int M[2][2])
{
int x = F[0][0]*M[0][0] + F[0][1]*M[1][0];
int y = F[0][0]*M[0][1] + F[0][1]*M[1][1];
int z = F[1][0]*M[0][0] + F[1][1]*M[1][0];
int w = F[1][0]*M[0][1] + F[1][1]*M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
/* Driver program to test above function */
int main()
{
int n = 9;
printf("%d", fib(9));
getchar();
return 0;
}
//Fibonacci Series using Optimized Method
class fibonacci
{
/* function that returns nth Fibonacci number */
static int fib(int n)
{
int F[][] = new int[][]{{1,1},{1,0}};
if (n == 0)
return 0;
power(F, n-1);
return F[0][0];
}
static void multiply(int F[][], int M[][])
{
int x = F[0][0]*M[0][0] + F[0][1]*M[1][0];
int y = F[0][0]*M[0][1] + F[0][1]*M[1][1];
int z = F[1][0]*M[0][0] + F[1][1]*M[1][0];
int w = F[1][0]*M[0][1] + F[1][1]*M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
/* Optimized version of power() in method 4 */
static void power(int F[][], int n)
{
if( n == 0 || n == 1)
return;
int M[][] = new int[][]{{1,1},{1,0}};
power(F, n/2);
multiply(F, F);
if (n%2 != 0)
multiply(F, M);
}
/* Driver program to test above function */
public static void main (String args[])
{
int n = 9;
System.out.println(fib(n));
}
}
/* This code is contributed by Rajat Mishra */
# Fibonacci Series using
# Optimized Method
# function that returns nth
# Fibonacci number
def fib(n):
F = [[1, 1],
[1, 0]]
if (n == 0):
return 0
power(F, n - 1)
return F[0][0]
def multiply(F, M):
x = (F[0][0] * M[0][0] +
F[0][1] * M[1][0])
y = (F[0][0] * M[0][1] +
F[0][1] * M[1][1])
z = (F[1][0] * M[0][0] +
F[1][1] * M[1][0])
w = (F[1][0] * M[0][1] +
F[1][1] * M[1][1])
F[0][0] = x
F[0][1] = y
F[1][0] = z
F[1][1] = w
# Optimized version of
# power() in method 4
def power(F, n):
if( n == 0 or n == 1):
return;
M = [[1, 1],
[1, 0]];
power(F, n // 2)
multiply(F, F)
if (n % 2 != 0):
multiply(F, M)
# Driver Code
if __name__ == "__main__":
n = 9
print(fib(n))
# This code is contributed
# by ChitraNayal
// Fibonacci Series using
// Optimized Method
using System;
class GFG
{
/* function that returns
nth Fibonacci number */
static int fib(int n)
{
int[,] F = new int[,]{{1, 1},
{1, 0}};
if (n == 0)
return 0;
power(F, n - 1);
return F[0, 0];
}
static void multiply(int[,] F,
int[,] M)
{
int x = F[0, 0] * M[0, 0] +
F[0, 1] * M[1, 0];
int y = F[0, 0] * M[0, 1] +
F[0, 1] * M[1, 1];
int z = F[1, 0] * M[0, 0] +
F[1, 1] * M[1, 0];
int w = F[1, 0] * M[0, 1] +
F[1, 1] * M[1, 1];
F[0, 0] = x;
F[0, 1] = y;
F[1, 0] = z;
F[1, 1] = w;
}
/* Optimized version of
power() in method 4 */
static void power(int[,] F, int n)
{
if( n == 0 || n == 1)
return;
int[,] M = new int[,]{{1, 1},
{1, 0}};
power(F, n / 2);
multiply(F, F);
if (n % 2 != 0)
multiply(F, M);
}
// Driver Code
public static void Main ()
{
int n = 9;
Console.Write(fib(n));
}
}
// This code is contributed
// by ChitraNayal
<script>
// Fibonacci Series using Optimized Method
// Function that returns nth Fibonacci number
function fib(n)
{
var F = [ [ 1, 1 ], [ 1, 0 ] ];
if (n == 0)
return 0;
power(F, n - 1);
return F[0][0];
}
function multiply(F, M)
{
var x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
var y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
var z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
var w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
F[0][0] = x;
F[0][1] = y;
F[1][0] = z;
F[1][1] = w;
}
// Optimized version of power() in method 4 */
function power(F, n)
{
if (n == 0 || n == 1)
return;
var M = [ [ 1, 1 ], [ 1, 0 ] ];
power(F, n / 2);
multiply(F, F);
if (n % 2 != 0)
multiply(F, M);
}
// Driver code
var n = 9;
document.write(fib(n));
// This code is contributed by gauravrajput1
</script>
Выход:
Временная Сложность: O(Логин)
Дополнительное Пространство: O(Logn), если мы учитываем размер стека вызовов функций, в противном случае O(1).
Способ 6 (O(Log n) Время)
Ниже приведена еще одна интересная рекуррентная формула, которую можно использовать для нахождения n-го числа Фибоначчи за O(Log n) времени.
If n is even then k = n/2:
F(n) = [2*F(k-1) + F(k)]*F(k)
If n is odd then k = (n + 1)/2
F(n) = F(k)*F(k) + F(k-1)*F(k-1)Как работает эта формула?
Формула может быть получена из приведенного выше матричного уравнения.
Принимая определитель с обеих сторон, мы получаем
(-1)n = Fn+1Fn-1 - Fn2
Более того, поскольку AnAm = An+m для любой квадратной матрицы A,
могут быть получены следующие тождества (они получены
из двух разных коэффициентов матричного произведения)
FmFn + Fm-1Fn-1 = Fm+n-1 ---------------------------(1)
Поместив n = n+1 в уравнение(1),
FmFn+1 + Fm-1Fn = Fm+n --------------------------(2)
Введем m = n в уравнение(1).
F2n-1 = Fn2 + Fn-12
Включение m = n в уравнение(2)
F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Источник: Wiki) --------
( Поставив Fn+1 = Fn + Fn-1 )
Чтобы доказать формулу, нам просто нужно сделать следующее
Если n четное, мы можем положить k = n/2
Если n нечетно, мы можем положить k = (n+1)/2Ниже приведена реализация вышеуказанной идеи.
// C++ Program to find n'th fibonacci Number in
// with O(Log n) arithmetic operations
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1000;
// Create an array for memoization
int f[MAX] = {0};
// Returns n'th fibonacci number using table f[]
int fib(int n)
{
// Base cases
if (n == 0)
return 0;
if (n == 1 || n == 2)
return (f[n] = 1);
// If fib(n) is already computed
if (f[n])
return f[n];
int k = (n & 1)? (n+1)/2 : n/2;
// Applying above formula [Note value n&1 is 1
// if n is odd, else 0.
f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))
: (2*fib(k-1) + fib(k))*fib(k);
return f[n];
}
/* Driver program to test above function */
int main()
{
int n = 9;
printf("%d ", fib(n));
return 0;
}
// Java Program to find n'th fibonacci
// Number with O(Log n) arithmetic operations
import java.util.*;
class GFG {
static int MAX = 1000;
static int f[];
// Returns n'th fibonacci number using
// table f[]
public static int fib(int n)
{
// Base cases
if (n == 0)
return 0;
if (n == 1 || n == 2)
return (f[n] = 1);
// If fib(n) is already computed
if (f[n] != 0)
return f[n];
int k = (n & 1) == 1? (n + 1) / 2
: n / 2;
// Applying above formula [Note value
// n&1 is 1 if n is odd, else 0.
f[n] = (n & 1) == 1? (fib(k) * fib(k) +
fib(k - 1) * fib(k - 1))
: (2 * fib(k - 1) + fib(k))
* fib(k);
return f[n];
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 9;
f= new int[MAX];
System.out.println(fib(n));
}
}
// This code is contributed by Arnav Kr. Mandal.
# Python3 Program to find n'th fibonacci Number in
# with O(Log n) arithmetic operations
MAX = 1000
# Create an array for memoization
f = [0] * MAX
# Returns n'th fibonacci number using table f[]
def fib(n) :
# Base cases
if (n == 0) :
return 0
if (n == 1 or n == 2) :
f[n] = 1
return (f[n])
# If fib(n) is already computed
if (f[n]) :
return f[n]
if( n & 1) :
k = (n + 1) // 2
else :
k = n // 2
# Applying above formula [Note value n&1 is 1
# if n is odd, else 0.
if((n & 1) ) :
f[n] = (fib(k) * fib(k) + fib(k-1) * fib(k-1))
else :
f[n] = (2*fib(k-1) + fib(k))*fib(k)
return f[n]
# Driver code
n = 9
print(fib(n))
# This code is contributed by Nikita Tiwari.
// C# Program to find n'th
// fibonacci Number with
// O(Log n) arithmetic operations
using System;
class GFG
{
static int MAX = 1000;
static int[] f;
// Returns n'th fibonacci
// number using table f[]
public static int fib(int n)
{
// Base cases
if (n == 0)
return 0;
if (n == 1 || n == 2)
return (f[n] = 1);
// If fib(n) is already
// computed
if (f[n] != 0)
return f[n];
int k = (n & 1) == 1 ? (n + 1) / 2
: n / 2;
// Applying above formula
// [Note value n&1 is 1 if
// n is odd, else 0.
f[n] = (n & 1) == 1 ? (fib(k) * fib(k) +
fib(k - 1) * fib(k - 1))
: (2 * fib(k - 1) + fib(k)) *
fib(k);
return f[n];
}
// Driver Code
static void Main()
{
int n = 9;
f = new int[MAX];
Console.WriteLine(fib(n));
}
}
// This code is contributed by mits
<?php
// PHP Program to find n'th
// fibonacci Number in with
// O(Log n) arithmetic operations
$MAX = 1000;
// Returns n'th fibonacci
// number using table f[]
function fib($n)
{
global $MAX;
// Create an array for memoization
$f = array_fill(0, $MAX, NULL);
// Base cases
if ($n == 0)
return 0;
if ($n == 1 || $n == 2)
return ($f[$n] = 1);
// If fib(n) is already computed
if ($f[$n])
return $f[$n];
$k = ($n & 1) ? ($n + 1) / 2 : $n / 2;
// Applying above formula
// [Note value n&1 is 1 if
// n is odd, else 0.
$f[$n] = ($n & 1) ? (fib($k) * fib($k) +
fib($k - 1) * fib($k - 1)) :
(2 * fib($k - 1) + fib($k)) * fib($k);
return $f[$n];
}
// Driver Code
$n = 9;
echo fib($n);
// This code is contributed
// by ChitraNayal
?>
<script>
// JavaScript Program to find n'th fibonacci Number in
// with O(Log n) arithmetic operations
const MAX = 1000;
// Create an array for memoization
var f = [...Array(MAX)];
f.fill(0);
// Returns n'th fibonacci number using table f[]
function fib(n) {
// Base cases
if (n == 0) return 0;
if (n == 1 || n == 2) return (f[n] = 1);
// If fib(n) is already computed
if (f[n]) return f[n];
var k = n & 1 ? (n + 1) / 2 : n / 2;
// Applying above formula [Note value n&1 is 1
// if n is odd, else 0.
f[n] =
n & 1
? fib(k) * fib(k) + fib(k - 1) * fib(k - 1)
: (2 * fib(k - 1) + fib(k)) * fib(k);
return f[n];
}
/* Driver program to test above function */
var n = 9;
document.write(fib(n));
// This code is contributed by rdtank.
</script>
Выход:
34Временная сложность этого решения равна O(Log n), так как мы разделяем проблему на две части в каждом рекурсивном вызове.
Способ 7
Другой подход(с использованием формулы) :
В этом методе мы непосредственно реализуем формулу для n-го члена в ряду Фибоначчи.
Fn = {[(√5 + 1)/2] ^ n} / √5
// C++ Program to find n'th fibonacci Number
#include<iostream>
#include<cmath>
int fib(int n) {
double phi = (1 + sqrt(5)) / 2;
return round(pow(phi, n) / sqrt(5));
}
// Driver Code
int main ()
{
int n = 9;
std::cout << fib(n) << std::endl;
return 0;
}
//This code is contributed by Lokesh Mohanty.
// C Program to find n'th fibonacci Number
#include<stdio.h>
#include<math.h>
int fib(int n) {
double phi = (1 + sqrt(5)) / 2;
return round(pow(phi, n) / sqrt(5));
}
int main ()
{
int n = 9;
printf("%d", fib(n));
return 0;
}
// Java Program to find n'th fibonacci Number
import java.util.*;
class GFG {
static int fib(int n) {
double phi = (1 + Math.sqrt(5)) / 2;
return (int) Math.round(Math.pow(phi, n)
/ Math.sqrt(5));
}
// Driver Code
public static void main(String[] args) {
int n = 9;
System.out.println(fib(n));
}
}
// This code is contributed by PrinciRaj1992
# Python3 program to find n'th
# fibonacci Number
import math
def fibo(n):
phi = (1 + math.sqrt(5)) / 2
return round(pow(phi, n) / math.sqrt(5))
# Driver code
if __name__ == '__main__':
n = 9
print(fibo(n))
# This code is contributed by prasun_parate
// C# Program to find n'th fibonacci Number
using System;
public class GFG
{
static int fib(int n)
{
double phi = (1 + Math.Sqrt(5)) / 2;
return (int) Math.Round(Math.Pow(phi, n)
/ Math.Sqrt(5));
}
// Driver code
public static void Main()
{
int n = 9;
Console.WriteLine(fib(n));
}
}
// This code is contributed by 29AjayKumar
<?php
// PHP Program to find n'th
// fibonacci Number
function fib($n)
{
$phi = (1 + sqrt(5)) / 2;
return round(pow($phi, $n) / sqrt(5));
}
// Driver Code
$n = 9;
echo fib($n) ;
// This code is contributed by Ryuga
?>
<script>
// Javascript Program to find n'th fibonacci Number
function fib(n) {
let phi = (1 + Math.sqrt(5)) / 2;
return Math.round(Math.pow(phi, n) / Math.sqrt(5));
}
let n = 9;
document.write(fib(n));
// This code is contributed by mukesh07.
</script>
Выход:
34Временная сложность: O(logn), это связано с тем, что для вычисления phi^n требуется
логарифмическая сложность временного пространства: O(1)
Способ 8
DP с использованием запоминания(подход сверху вниз)
Мы можем избежать повторной работы, проделанной методом 1, сохранив числа Фибоначчи, вычисленные до сих пор. Нам просто нужно сохранить все значения в массиве.
#include <bits/stdc++.h>
using namespace std;
int dp[10];
int fib(int n)
{
if (n <= 1)
return n;
// temporary variables to store
// values of fib(n-1) & fib(n-2)
int first, second;
if (dp[n - 1] != -1)
first = dp[n - 1];
else
first = fib(n - 1);
if (dp[n - 2] != -1)
second = dp[n - 2];
else
second = fib(n - 2);
// memoization
return dp[n] = first + second;
}
// Driver Code
int main()
{
int n = 9;
memset(dp, -1, sizeof(dp));
cout << fib(n);
getchar();
return 0;
// This code is contributed by Bhavneet Singh
}
import java.util.*;
class GFG{
// Initialize array of dp
static int[] dp = new int[10];
static int fib(int n)
{
if (n <= 1)
return n;
// Temporary variables to store
// values of fib(n-1) & fib(n-2)
int first, second;
if (dp[n - 1] != -1)
first = dp[n - 1];
else
first = fib(n - 1);
if (dp[n - 2] != -1)
second = dp[n - 2];
else
second = fib(n - 2);
// Memoization
return dp[n] = first + second;
}
// Driver Code
public static void main(String[] args)
{
int n = 9;
Arrays.fill(dp, -1);
System.out.print(fib(n));
}
}
// This code is contributed by sujitmeshram
# Initialize array of dp
dp = [-1 for i in range(10)]
def fib(n):
if (n <= 1):
return n;
global dp;
# Temporary variables to store
# values of fib(n-1) & fib(n-2)
first = 0;
second = 0;
if (dp[n - 1] != -1):
first = dp[n - 1];
else:
first = fib(n - 1);
if (dp[n - 2] != -1):
second = dp[n - 2];
else:
second = fib(n - 2);
dp[n] = first + second;
# Memoization
return dp[n] ;
# Driver Code
if __name__ == '__main__':
n = 9;
print(fib(n));
# This code contributed by Rajput-Ji
using System;
class GFG {
// Initialize array of dp
static int[] dp = new int[10];
static int fib(int n)
{
if (n <= 1)
return n;
// Temporary variables to store
// values of fib(n-1) & fib(n-2)
int first, second;
if (dp[n - 1] != -1)
first = dp[n - 1];
else
first = fib(n - 1);
if (dp[n - 2] != -1)
second = dp[n - 2];
else
second = fib(n - 2);
// Memoization
return dp[n] = first + second;
}
// Driver code
static void Main()
{
int n = 9;
Array.Fill(dp, -1);
Console.Write(fib(n));
}
}
// This code is contributed by divyeshrabadiya07.
<script>
// Initialize array of dp
dp = Array.from({length: 10}, (_, i) => -1);
function fib(n)
{
if (n <= 1)
return n;
// Temporary variables to store
// values of fib(n-1) & fib(n-2)
var first, second;
if (dp[n - 1] != -1)
first = dp[n - 1];
else
first = fib(n - 1);
if (dp[n - 2] != -1)
second = dp[n - 2];
else
second = fib(n - 2);
// Memoization
return dp[n] = first + second;
}
// Driver Code
var n = 9;
document.write(fib(n));
// This code is contributed by Amit Katiyar
</script>
Выход:
34