Donat un número n , imprimir n-è número de Fibonacci .
int a la cadena
Els nombres de Fibonacci són els nombres de la següent seqüència d'enters: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..
Exemples:
Problema recomanat per resoldre el problema [/Tex] amb valors de llavors iEntrada: n = 1
Sortida: 1
Entrada: n = 9
Sortida: 34
font de làtexEntrada: n = 10
Sortida: 55
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. 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> |
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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;> }> |
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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> |
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Python 3
# 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> |
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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.> |
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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> > |
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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. ?>> |
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Sortida
34>
Complexitat temporal: O(n)
Espai auxiliar: O(1)
Enfocament de recursivitat per trobar i imprimir números enèsims de Fibonacci:
En termes matemàtics, la seqüència Fn de nombres de Fibonacci es defineix per la relació de recurrència:
amb valors de llavors i
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El nombre Nè de Fibonacci es pot trobar utilitzant la relació de recurrència mostrada a dalt:
- si n = 0, després retorna 0.
- Si n = 1, hauria de retornar 1.
- Per a n> 1, hauria de retornar Fn-1+ Fn-2
A continuació es mostra la implementació de l'enfocament anterior:
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> |
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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;> }> |
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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 */> |
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Python 3
# 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.> |
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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> |
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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));> |
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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 ?>> |
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Sortida
34>
Complexitat temporal: exponencial, ja que cada funció anomena altres dues funcions.
Complexitat de l'espai auxiliar: O(n), ja que la profunditat màxima de l'arbre de recursivitat és n.
Enfocament de programació dinàmica per trobar i imprimir números enèsims de Fibonacci:
Considereu l'arbre de recursivitat per al 5è nombre de Fibonacci de l'enfocament anterior:
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)>
Si veieu, la mateixa trucada de mètode s'està fent diverses vegades per al mateix valor. Això es pot optimitzar amb l'ajuda de la programació dinàmica. Podem evitar el treball repetit realitzat en l'enfocament de la recursència emmagatzemant els nombres de Fibonacci calculats fins ara.
Enfocament de programació dinàmica per trobar i imprimir números enèsims de Fibonacci:
A continuació es mostra la implementació de l'enfocament anterior:
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> |
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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;> }> |
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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 */> |
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Python 3
# 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>))> |
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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.> |
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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> > > |
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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. ?>> |
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Sortida
34>
Complexitat temporal : O(n) per a n donat
Espai auxiliar : O(n)
alfabet als números
Aproximació a la potència enèima de la matriu per trobar i imprimir els números enèsims de Fibonacci
Aquest enfocament es basa en el fet que si multipliquem n vegades la matriu M = {{1,1},{1,0}} per si mateixa (és a dir, calculem la potència (M, n)), aleshores obtenim la (n +1)è nombre de Fibonacci com a element de fila i columna (0, 0) de la matriu resultant.
- Si n és parell aleshores k = n/2:
- Per tant, enèsimo nombre de Fibonacci = F(n) = [2*F(k-1) + F(k)]*F(k)
- Si n és senar aleshores k = (n + 1)/2:
- Per tant, enèsimo nombre de Fibonacci = F(n) = F(k)*F(k) + F(k-1)*F(k-1)
Com funciona aquesta fórmula?
La fórmula es pot derivar de l'equació matricial.
Prenent determinant per ambdós costats, obtenim (-1)n= Fn+1Fn-1– Fn2
gimp guardant com a jpegA més, des que AnAm= An+mper a qualsevol matriu quadrada A, es poden derivar les identitats següents (s'obtenen a partir de dos coeficients diferents del producte de la matriu)
FmFn+ Fm-1Fn-1= Fm+n-1 —————————(1)
Posant n = n+1 a l'equació (1),
FmFn+1+ Fm-1Fn= Fm+n —————————(2)
comanda arp-aPosant m = n a l'equació (1).
F2n-1= Fn2+ Fn-12
Posant m = n a l'equació (2)
F2n= (Fn-1+ Fn+1)Fn= (2Fn-1+ Fn)Fn———
(Posant Fn+1 = Fn + Fn-1)
Per aconseguir que la fórmula es demostri, només hem de fer el següent
- Si n és parell, podem posar k = n/2
- Si n és senar, podem posar k = (n+1)/2
A continuació es mostra la implementació de l'enfocament anterior
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> |
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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;> }> |
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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 */> |
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Python 3
# 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> |
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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> |
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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> > |
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Sortida
34>
Complexitat temporal: O (Registre n)
Espai auxiliar: O(Log n) si considerem la mida de la pila de trucades de funció, en cas contrari O(1).
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