只有50分，大佬求解

LJC13618059782 @ 2023-10-30 21:38:24

#include<bits/stdc++.h>
using namespace std;
long long a[40];//全局变量自动初始化为0 
int f(int x)//f函数的作用走到第x级台阶的走法总数。 
{
    if(x==1)return 1;
    if(x==2)return 2;

    if(a[x-1]==-1)a[x-1]=f(x-1);
    if(a[x-2]==-1)a[x-2]=f(x-2);
    return a[x-1]+a[x-2];
}

int main()
{
    long long n,ans; 
    while(cin>>n)
    {
        for(int i=1;i<=n;++i)a[i]=-1;//将a数组初始化为-1
        ans=f(n);
        cout<<ans<<endl;
    }
    return 0;
}
//只有50分，大佬求解

by Night_fall @ 2023-10-30 21:40:30

要用高精度，单纯递推位数不够

by Syncc @ 2023-10-30 21:41:31

需要高精啊，不过用 unsigned long long 能60分

by LJC13618059782 @ 2023-10-30 21:45:38

@Razer_System 知道了 谢谢

by LJC13618059782 @ 2023-10-30 21:46:22

@Night_fall 知道了 谢谢

by LJC13618059782 @ 2023-10-30 21:57:02

#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> add(vector<int> &A, vector<int> &B)
{
    if (A.size() < B.size()) return add(B, A);
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    {
        t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }

    if (t) C.push_back(t);
    return C;
}
int main(){
    while(cin>>n){
        vector<int> a={1},b={2},c;
        if(n==1){
            cout<<1<<endl;
        }
        if(n==2){
            cout<<2<<endl;
        }
        if(n>=3){
            for(int i=3;i<=n;i++){
                c=add(a,b);
                a=b;
                b=c;
            }
            for (int i = c.size() - 1; i >= 0; i--)
            cout<<c[i];
        cout<<endl;
        }      
    }
    return 0;

}
//谢谢，我知道了