小黄鸭

二分水下高度为$hh$，因为$\rho vg=mg,\rho =1\rho vg\; =\; mg,\rho \; =\; 1$,所以$m=vm\; =\; v$，所以我们可以求水下的体积来得到$hh$，从而得到需要的答案$2\ast r-h2*r-h\backslash \backslash$. 球方程为：$x^2+(y-r{)}^2=r^{2}x^2\; +\; (y-r)^2\; =\; r^2$,水下面积为$\pi r^2\backslash pi\; r^2$，将水面下方的物体垂直的分为一片一片的小圆柱体，然后积分区间就为[0,h],即${\int }_0^h(r^2-(y-r{)}^2)\pi \mathrm{d}y\backslash int\_\{0\}^\{h\}(r^2-(y-r)^2)\backslash pi\backslash mathrm\{d\}y$，最后久可以得到$V=r\ast h^2-\frac{h^3}{3}\pi V\; =\; r*h^2\; -\; \backslash frac\{h^3\}\{3\}\; \backslash pi$

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
# include<bits/stdc++.h>
# define  PII pair<int,int>
# define  PDI pair<double,int>
# define  PIII pair<int,PII>
# define int long long
using namespace std;
const int N = 2e5+10;
const int mod = 32768;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1); 

int T,n,m,k1,k2,ans,k;

void solve(){
	      double r,m;
	      cin >> r >> m;
	      double L = 0 , R = 2 * r ;
	      while(R - L >= 1e-12){
		  	  double mid = (L + R )/2;
		  	  if(pi * (r*mid*mid-mid*mid*mid/3) > m) R = mid;
		  	   else L = mid;
		  }
		  cout<<fixed<<setprecision(2)<<2*r-L;
	       
}
/*
*/
signed main(){
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
   // cin >> T;
     T = 1;
    while(T--){
		 solve();
	}
 	return 0;
}