N-DE FIBONACCI-NUMMER - TECHCODEVIEW.COM

Gegeven een nummer N , afdrukken n-de Fibonacci-getal .

De Fibonacci-getallen zijn de getallen in de volgende gehele reeks: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..

Voorbeelden:

Invoer : n = 1

Uitgang: 1
wanneer kwam windows 7 uit

Invoer : n = 9

Uitgang: 3. 4

Invoer : n = 10

Uitgang: 55

Aanbevolen probleem Probleem oplossen [/Tex]met zaadwaarden en F_0 = 0

C++

// Fibonacci Series using Space Optimized Method> #include> 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>

C

// Fibonacci Series using Space Optimized Method> #include> 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

// Java program for Fibonacci Series using Space> // Optimized Method> public> 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>

Python3

# 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#

// 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.>

Javascript

> // 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> >

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. ?>>

Uitvoer

34>

Tijdcomplexiteit: Op)
Hulpruimte: O(1)

Recursiebenadering voor het vinden en afdrukken van N-de Fibonacci-getallen:

In wiskundige termen wordt de reeks Fn van Fibonacci-getallen gedefinieerd door de herhalingsrelatie: $F_{n} = F_{n-1} + F_{n-2}$ met zaadwaarden en En .

Het N-de Fibonacci-getal kan worden gevonden met behulp van de hierboven weergegeven herhalingsrelatie:

als N = 0, en retourneer dan 0.
Als n = 1, dan zou het 1 moeten retourneren.
Voor n> 1 zou het F moeten retourneren_n-1+ F_n-2

Hieronder vindt u de implementatie van de bovenstaande aanpak:

C++

// Fibonacci Series using Recursion> #include> 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 << n <<>'th Fibonacci Number: '> << fib(n);> >return> 0;> }> // This code is contributed> // by Akanksha Rai>

C

// Fibonacci Series using Recursion> #include> int> fib(>int> n)> {> >if> (n <= 1)> >return> n;> >return> fib(n - 1) + fib(n - 2);> }> int> main()> {> >int> n = 9;> >printf>(>'%dth Fibonacci Number: %d'>, n, fib(n));> >return> 0;> }>

Java

// Fibonacci Series using Recursion> import> java.io.*;> 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(> >n +>'th Fibonacci Number: '> + fib(n));> >}> }> /* This code is contributed by Rajat Mishra */>

Python3

# Fibonacci series using recursion> def> fibonacci(n):> >if> n <>=> 1>:> >return> n> >return> fibonacci(n>->1>)>+> fibonacci(n>->2>)> if> __name__>=>=> '__main__'>:> >n>=> 9> >print>(n,>'th Fibonacci Number: '>)> >print>(fibonacci(n))> ># This code is contributed by Manan Tyagi.>

C#

// 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(n +>'th Fibonacci Number: '> + Fib(n));> >}> }> // This code is contributed by Sam007>

Javascript

// Javascript program for Fibonacci Series> // using Recursion> function> Fib(n) {> >if> (n <= 1) {> >return> n;> >}>else> {> >return> Fib(n - 1) + Fib(n - 2);> >}> }> // driver code> let n = 9;> console.log(n +>'th Fibonacci Number: '> + Fib(n));>

PHP

 // PHP program for Fibonacci Series // using Recursion function Fib($n) {  if ($n <= 1) {  return $n;  }  else {  return Fib($n - 1) + Fib($n - 2);  } } // driver code $n = 9; echo $n . 'th Fibonacci Number: ' . Fib($n); // This code is contributed by Sam007 ?>>

Uitvoer

34>

Tijdcomplexiteit: exponentieel, aangezien elke functie twee andere functies aanroept.
Complexiteit van de hulpruimte: O(n), aangezien de maximale diepte van de recursieboom n is.

Dynamische programmeerbenadering voor het vinden en afdrukken van N-de Fibonacci-getallen:

Beschouw de recursieboom voor het 5e Fibonacci-getal uit de bovenstaande benadering:

 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)>

Als u ziet, wordt dezelfde methodeaanroep meerdere keren uitgevoerd voor dezelfde waarde. Dit kan worden geoptimaliseerd met behulp van Dynamisch Programmeren. We kunnen het herhaalde werk van de recursiebenadering vermijden door de tot nu toe berekende Fibonacci-getallen op te slaan.

Dynamische programmeerbenadering voor het vinden en afdrukken van N-de Fibonacci-getallen:

Hieronder vindt u de implementatie van de bovenstaande aanpak:

C++

// C++ program for Fibonacci Series> // using Dynamic Programming> #include> 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>

C

// Fibonacci Series using Dynamic Programming> #include> 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;> }>

Java

// Fibonacci Series using Dynamic Programming> public> 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 */>

Python3

# 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#

// 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.>

Javascript

> // 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> > >

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.  ?>>

Uitvoer

34>

Tijdcomplexiteit : O(n) voor gegeven n
Hulpruimte : Op)

N-de kracht van matrixbenadering voor het vinden en afdrukken van N-de Fibonacci-getallen

Deze benadering is gebaseerd op het feit dat als we n keer de matrix M = {{1,1},{1,0}} met zichzelf vermenigvuldigen (met andere woorden het vermogen (M, n) berekenen), we de (n +1)e Fibonacci-getal als het element in rij en kolom (0, 0) in de resulterende matrix.

Als n even is, dan is k = n/2:
Daarom N-de Fibonacci-getal = F(n) = [2*F(k-1) + F(k)]*F(k)
Als n oneven is, dan is k = (n + 1)/2:
Daarom N-de Fibonacci-getal = F(n) = F(k)*F(k) + F(k-1)*F(k-1)

Hoe werkt deze formule?
De formule kan worden afgeleid uit de matrixvergelijking.

$egin{bmatrix}1 & 1 1 & 0 end{bmatrix}^n = egin{bmatrix}F_{n+1} & F_n F_n & F_{n-1} end{bmatrix}$

wat is de grootte van mijn monitorscherm

Als we aan beide kanten de determinant nemen, krijgen we (-1)^N= F_n+1F_n-1- F_N²

Bovendien, aangezien A^NA^M= EEN^n+mvoor elke vierkante matrix A kunnen de volgende identiteiten worden afgeleid (ze worden verkregen uit twee verschillende coëfficiënten van het matrixproduct)

F_MF_N+ F_m-1F_n-1= F_{m+n-1 —————————(1)}

Door n = n+1 in vergelijking(1) te plaatsen,

F_MF_n+1+ F_m-1F_N= F_{m+n ————————–(2)}

Door m = n in vergelijking(1) te plaatsen.

F_2n-1= F_N²+ F_n-1²

M = n in vergelijking(2) plaatsen

F_2n= (F_n-1+ F_n+1)F_N= (2F_n-1+ F_N)F_N——–

(Door Fn+1 = Fn + Fn-1 te plaatsen)

Om de formule te bewijzen, hoeven we alleen maar het volgende te doen

Als n even is, kunnen we k = n/2 stellen
Als n oneven is, kunnen we k = (n+1)/2 stellen

Hieronder vindt u de implementatie van bovenstaande aanpak

C++

// Fibonacci Series using Optimized Method> #include> 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>

C

#include> 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;> }>

Java

// Fibonacci Series using Optimized Method> public> 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 */>

Python3

# 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>

C#

// 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>

Javascript

> // 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> // Driver code> var> n = 9;> document.write(fib(n));> // This code is contributed by gauravrajput1> >

Uitvoer

34>

Tijdcomplexiteit: O(Logboek n)
Hulpruimte: O(Log n) als we rekening houden met de functieaanroepstapelgrootte, anders O(1).

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Grote Fibonacci-getallen op Java