/***********
P3382 【模板】三分法
***********/
#include <bits/stdc++.h>
using namespace std;
const int N = 20;
double a[N];
int n;
double f(double x) //此处可以用秦九韶算法简化为O(n)的!
{
double ans = 0;
for(int i=1; i<=n+1; i++)
{
ans += 1.0 * pow(x,n - i + 1) * a[i];
}
return ans;
}
int main()
{
double l,r;
cin >> n >> l >> r;
for(int i=1; i<=n+1; i++) cin >> a[i];
double ans = 0;
while(r - l > 1e-7)
{
double lmid = l + (r - l) / 3,rmid = r - (r - l) / 3;
if(f(lmid) >= f(rmid)) r = rmid,ans = rmid;
else l = lmid;
}
printf("%.6f\n",ans);
return 0;
}