lq2.3フィボナッチ数列

//多个小规模子问题 

#include <stdio.h>

int feibo(int x);

int main()
{
    
    
	int x;
	for(;;)
	{
    
    
		printf("请输入第x位:");
	scanf("%d",&x);
	printf("第%d位为%d\n",x,feibo(x)); 	
	}
} 

int feibo(int x)
{
    
    
	if((x==1)|(x==2)) 
		return 1;
	return feibo(x-1)+feibo(x-2);
}

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転載: blog.csdn.net/June159/article/details/108453826