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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?

实现1：

int f1(int x) {timer1++;return (x ==0||x==1) ? 1 : f1(x-1)+f1(x-2);}

实现2：

int dp[50];

int f(int x) {

     shen++; 

     if(x == 0 || x == 1 ){ dp[0]  = 1;dp[1]  = 1; return 1;}

     if(dp[x] != -1)  return dp[x];

     return dp[x]= f(x - 2)+f( x - 1);

}