Step-up problem (recursive design)

Question: There are n steps in the staircase. One, two, or three steps can be taken in one step. How many ways to go up the stairs?

Code:

#include<iostream>
using namespace std;

int cnt=0;//用于计数
void f(int n)//n表示还剩你个台阶需要走
{
    if(n<0) return; //防止死循环
    if(n==0)
    {
        cnt++;
        return;
    }
    f(n-1); //走一步
    f(n-2); //走两步
    f(n-3); //走三步
}
int main()
{
    int n;
    cin>>n;
    f(n);
    cout<<cnt;
    return 0;
}

Defects: When n is relatively large, it takes a lot of time and may even burst the stack. Analysis shows that there are a lot of repetitive sub-problems in the above recursion, and a lot of useless work is done. In order to solve the repetitive sub-problem, memory recursion can be used for optimization.

Code:

#include<iostream>
using namespace std;


int dp[1001];//假设楼梯的台阶数可以达到1000
int f(int n)//n表示还剩你个台阶需要走
{
    if(dp[n]!=0) return dp[n];
    int cnt=0;//用于计数
    if(n<0) return 0; //防止死循环
    if(n==0)
    {
        cnt++;
        return cnt;
    }
    cnt+=f(n-1); //走一步
    cnt+=f(n-2); //走两步
    cnt+=f(n-3); //走三步
    dp[n]=cnt;
    return cnt;
}
int main()
{
    int n;
    cin>>n;
    cout<<f(n);
    return 0;
}