sui13127198996 @ 2023-08-22 19:05:02
#include<bits/stdc++.h>
using namespace std;
long long f[5001];
int main(){
int n;
cin>>n;
f[1]=1;
f[2]=2;
for(int i=3;i<=n;i++){
f[i]=f[i-1]+f[i-2];
}
cout<<f[n];
return 0;
}
求大佬帮助,非常感谢
by _Anonymous_ @ 2023-08-22 19:09:46
@sui13127198996
高精度,n的范围到5000
by sui13127198996 @ 2023-08-22 19:10:29
@Anonymous 感谢,已关注