ans=1.0*(pow((1+sqrt(5))/2,n)-pow((1-sqrt(5))/2,n))/sqrt((5));
AC代码:
#include <bits/stdc++.h> using namespace std; int n; double ans; int main() { cin>>n; ans=1.0*(pow((1+sqrt(5))/2,n)-pow((1-sqrt(5))/2,n))/sqrt((5)); printf("%.2f",ans); }
By signing up a HydroOJ universal account, you can submit code and join discussions in all online judging services provided by us.
Using your HydroOJ universal account