大佬995,5只得50分！！！！

_Maverick_ @ 2023-10-06 08:40:55

#include<iostream>
using namespace std;
int f[21];
int main(){
    int n;
    cin>>n;
    f[0]=1;
    f[1]=1;
    for(int i=2;i<=n;i++){
    f[i]=f[i-1]+f[i-2];
}
    cout<<f[n];
    return 0;
}

by jqQt0220 @ 2023-10-06 08:42:03

要用高精

by kbzcz @ 2023-10-06 08:42:07

你看看数据范围

by ListentoRain @ 2023-10-06 08:42:25

《对于60%的数据，N<=50》

by _YTY_ @ 2023-10-06 08:42:42

@myh00198100,此题需要高精度

by _Maverick_ @ 2023-10-06 08:44:35

@YTY 哦哦

by cspwhp @ 2023-10-06 09:18:25

《n<=50》

by zhouyk0501 @ 2023-10-08 15:40:58

#include<bits/stdc++.h>
using namespace std;
string dp[10000];
string r(string s){
    string s1="";
    for(int i=s.size()-1;i>=0;i--){
        s1+=s[i];
    }return s1;
}
string add(string s,string s1){
    int t=0;
    while(s.size()<s1.size()) s='0'+s;
    while(s1.size()<s.size()) s1='0'+s1;
//  cout<<s<<" "<<s1;
    for(int i=s.size()-1;i>=0;i--){
        int a=s[i]-'0',b=s1[i]-'0';
        s[i]=(a+b+t)%10+'0';
//      cout<<s2;
        t=(a+b+t)/10;
//      cout<<s2<<"\n";
    }//cout<<s2;
    if(t>0){
        string s3=to_string(t);
        s3=r(s3);
        s=r(s);
        s+=s3;
        s=r(s);
    }return s;
}
int main(){
    dp[1]="1";
    dp[2]="2";
    for(int i=3;i<=5010;i++) dp[i]=add(dp[i-1],dp[i-2]);
    int n;
    cin>>n;
    cout<<dp[n];
}