c语言实例之腾讯面试题_return tencent(n - 1) + tencent(n - 2)

<span style="font-size:18px;">#include <stdio.h>
#include <stdlib.h>
 
//腾讯面试问题:有50个台阶，一次走一步或者走两步，有多少种可能
 
double tencent(int n)
{
	if(n==1)
	{
		return 1.0;
	}
	else if(n==2)
	{
		return 2.0;
	}
	else
	{
		return tencent(n-1)+tencent(n-2);
	}	
 
}
 
double tt(int n)
{
	double n1 = 1.0;
	double n2 = 2.0;
	double n3 = 0;
	int i;
	for( i=0;i < n;i++)
	{
		n3 = n1 + n2;
		n1 = n2;
		n2 = n3;
	}
	return n3;
 
}
double ttt(int n)
{
	int i;
	double num[50];
	num[0] = 1.0;
	num[1] = 2.0;
	
	for(i=2;i<n;i++)
	{
		num[i] = num[i-1]+num[i-2];
	}
	return num[n-1];
 
}
 
 
int main1()
{
	printf("总步数为:%lf\n",tencent(12));
	printf("总步数为:%lf\n",tt(10));
	printf("总步数为:%lf\n",ttt(12));
	system("pause");
	return 0;
}</span>