再求f(x,n)
#include<iostream> #include<cmath> #include <bits/stdc++.h> using namespace std; double a(double n,double x){ if(n==0){ return 1;//结尾条件 }else{ return x/(n+a(n-1,x));//表达式 } } int main(){ double x,n,y; cin>>x>>n; y=a(n,x); cout<<fixed<<setprecision(2)<<y; return 0; }