第n个斐波那契数的C / C++程序

用数学术语来说，斐波纳契数的序列Fn由递归关系定义

Fn = Fn-1 + Fn-2

具有种子值

F0 = 0 and F1 = 1.

方法1(使用递归)

//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("%d", fib(n));
  getchar();
  return 0;
}

方法2(动态编程)

//Fibonacci Series using Dynamic Programming
#include
  
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;
}

方法3(带有空间优化的动态规划)

// 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;
}

方法4(分而治之)

#include 
  
/* 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;
}

方法5(分而治之)

#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;
}

请参阅有关斐波那契数字的程序的完整文章，以了解更多详细信息！

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