/*https://oj.uz/problem/view/CEOI11_bal
first, if pre ballon radius is R, positon is x, new radius is r, position is y
we got (R+r)^2= (R-r)^2 +(y-x)^2 => r= (y-x)^2/ 4R
sort by position, for each ballon, if it's radius bigger than pre ones, the pre ones will never impact future one
so, we use stack save all radius, for new one , compare all ones in stack, compare with them find the real R.
for each new R, if it's bigger than pre one, pop pre one.
*/
#include <bits/stdc++.h>
using namespace std;
double getr(double lastx,double lastr,double x){
return (x-lastx) * (x-lastx) / 4 / lastr;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
//freopen("hayfeast.in","r",stdin);
//freopen("hayfeast.out","w",stdout);
int n;
cin >> n ;
vector <double> ans(n);//save answer
stack<pair<double,double>> st;//first, x second, r
double x,r;//r is limitation for radius
for (int i = 0; i < n; i++) {
cin >> x >> r;
while (!st.empty()) {
double lastr = st.top().second;
double lastx = st.top().first;
double maxr = getr(lastx,lastr,x);
r = min(r,maxr);
if (r > lastr) {
//bigger than last, last is useless,next time will check pre pre one to find new r
st.pop();
}
else break;//smaller tha preone,no need to check others
}
st.push({x,r});
ans[i] = r;
}
cout << fixed << setprecision(3) ;
for (auto t:ans) cout << t << '\n';
return 0;
}
