50 Points of Jsoi R2d1t3

#include <bits/stdc++.h>

using namespace Std;

int n,r,x,y;

Double ans;

Double dis (int x,int y) {return sqrt (x*x+y*y);}

int main () {

cin>>n>>r;

for (int i=1;i<=n;i++) {Cin>>x>>y;ans=max (Ans,abs (Dis (x, y)-R);}

printf ("%.10lf", ans);

return 0;

}

Positive solution: two points + network flow
Residual product: The lower limit of the dichotomy.
The reasons for the above 50-point approach:
1.99.999999% unexpected mess of the wrong person.
2. Positive solution fried, two points + network flow of the routine fried, the output lower limit to be comforted.
3. For a large number of points away from the origin point near, a small number of points away from the origin of the super far from the data, and out of the data in order to get rid of some other pointers, the last 5 points have 3 such data.
Impact:
Negative: 1. Roughly none.
Positive: 2. Make certain minesweeper/spider Solitaire players lose their goals? (There is absolutely no irony in the way the day2ak after a big guy.) )
It is said that this problem can be all kinds of mess? Burst 0 Konjac Konjac beats me.
Yuan Fang, you * * *?

50 Points of Jsoi R2d1t3