用递归法求斐波那契数列的方法

#include<stdio.h>
int fib(int n)
{
	if (n ==1||n==2)
		return 1;
	else
		return fib(n - 1) + fib(n - 2);
}
int main()
{
	int n = 0;
	scanf("%d", &n);
	int a = 0;
	printf("前%d项斐波那契数列的值分别为：",n);
	for (int i = 1; i <= n; i++)
	{
		a = fib(i);
		printf("%d ", a);
	}
	
	printf("\n第%d项斐波那契数列的值为:%d\n", n,a);
	return 0;
}