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?
Note: Given n will be a positive integer.
Example 1:
Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps
Example 2:
Input: 3 Output: 3 Explanation: There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step
思路：
首先先求出最大有可以由几个2组成，假设最大可以有y个2，那么在有y个2的情况，我们可以有
Cny种排列方法。 以此类推。 一共可以有
Cn0+Cn1+Cn2+...Cny 种方法。
代码：
class Solution {
public:
int climbStairs(int n) {
int cnt = 1;
int temp = n;
if(n==0||n==1)
return 1;
int y=n/2;//最多y个2
int i=0;
int ans=0;
while(i<=y)
{
int x=n-i*2;//x个1;
int k=0;
double c=1;
int t=i;
while(k<i)
{
double temp=c;
c*=x+i-k;
if(c>INT_MAX)
{
if(t>0)
{
c=temp;
c/=t;
t--;
}
}
else
k++;
}
while(t>0)
{
c/=t;
t--;
}
ans+=c;
i++;
}
return ans;
}
};