题目描述：
You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
思路:斐波那契数列的应用。f(n)=f(n-1)+f(n-2)，其中f(1)=1,f(2)=2。此题如果用递归会超时，所以采用非递归的方法求斐波那契数列的第n项。
代码：
int Solution::climbStairs(int n)
{
int n1 = 1;
int n2 = 2;
if(n == 1)
return 1;
if(n == 2)
return 2;
while(n > 2)
{
int temp = n1;
n1 = n2;
n2 = n2 + temp;
n--;
}
return n2;
}原文：http://blog.csdn.net/yao_wust/article/details/40538109