求解斐波那契數列的第n個數

斐波那契數列： 形如 

1     2     3     5      8      13      21 ............

一、遞歸求解

#include <stdio.h>
#include <stdlib.h>
int Fibonacci(int n)
{
	int a = 1;
	int b = 2;
	if (n == 1)
	{
		return a;
	}
	else if (n == 2)
	{
		return b;
	}
	else
	{
		return  Fibonacci(n - 2) + Fibonacci(n - 1);
	}
}

int main()
{
	int n = 0;
	int ret = 0;
	scanf("%d", &n);

	ret = Fibonacci(n);
	printf("%d\n", ret);
	system("pause");
	return 0;
}

二、非遞歸方法求解

#include <stdio.h>
#include <stdlib.h>

int fibonacci(int n)
{
	int a = 1;
	int b = 2;
	int c = 0;
	if (n == 1 || n == 2)
	{
		return n;
	}
	else
	{
		while (n > 2)
		{
			c = a + b;
			a = b;
			b = c;
			n--;
		}
		return c;
	}
}
int main()
{
	int n = 0;
	int ret = 0;
	scanf("%d", &n);
	ret = fibonacci(n);
	printf("%d\n", ret);

	system("pause");
	return 0;
}

歡迎各位批評指正！