C/C++ Program for n-th Fibonacci number - GeeksforGeeks

20 Nov, 2018
In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
    Fn = Fn-1 + Fn-2
with seed values
   F0 = 0 and F1 = 1.
Method 1 ( Use recursion )
//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;}
Method 2 (Dynamic Programming)
//Fibonacci Series using Dynamic Programming#include<stdio.h>  int fib(int n){  /* Declare an array to store Fibonacci numbers. */  int f[n+1];  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;}
Method 3 (Dynamic Programming with Space Optimization)
// 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;}
Method 4 (Divide and Conquer)
#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;}
Method 5 (Divide and Conquer)
#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;}
Please refer complete article on Program for Fibonacci numbers for more details!
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