C Program for n-th Fibonacci number
Last Updated :
14 Sep, 2023
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 )
C
#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;
}
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Time Complexity: O(2n)
Auxiliary Space: O(n)
Method 2 (Dynamic Programming)
C
#include<stdio.h>
int fib( int n)
{
int f[n+1];
int i;
f[0] = 0;
f[1] = 1;
for (i = 2; i <= n; i++)
{
f[i] = f[i-1] + f[i-2];
}
return f[n];
}
int main ()
{
int n = 9;
printf ( "%d" , fib(n));
getchar ();
return 0;
}
|
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3 (Dynamic Programming with Space Optimization)
C
#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;
}
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Time Complexity: O(n)
Auxiliary Space: O(1)
Method 4 (Divide and Conquer)
C
#include <stdio.h>
void multiply( int F[2][2], int M[2][2]);
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}};
for (i = 2; i <= n; i++)
multiply(F, M);
}
int main()
{
int n = 9;
printf ( "%d" , fib(n));
getchar ();
return 0;
}
|
Method 5 (Divide and Conquer)
C
#include <stdio.h>
void multiply( int F[2][2], int M[2][2]);
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 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;
}
int main()
{
int n = 9;
printf ( "%d" , fib(9));
getchar ();
return 0;
}
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Please refer complete article on Program for Fibonacci numbers for more details!
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