题目

The New Vasjuki village is stretched along the motorway and that's why every house on it is characterized by its shift relative to some fixed point — the x_i coordinate. The village consists of n houses, the i-th house is located in the point with coordinates of x_i.

TELE3, a cellular communication provider planned to locate three base stations so as to provide every house in the village with cellular communication. The base station having power d located in the point t provides with communication all the houses on the segment [t - d, t + d] (including boundaries).

To simplify the integration (and simply not to mix anything up) all the three stations are planned to possess the equal power of d. Which minimal value of d is enough to provide all the houses in the village with cellular communication.

Input

The first line contains an integer n (1 ≤ n ≤ 2·10⁵) which represents the number of houses in the village. The second line contains the coordinates of houses — the sequence x₁, x₂, ..., x_n of integer numbers (1 ≤ x_i ≤ 10⁹). It is possible that two or more houses are located on one point. The coordinates are given in a arbitrary order.

Output

Print the required minimal power d. In the second line print three numbers — the possible coordinates of the base stations' location. Print the coordinates with 6 digits after the decimal point. The positions of the stations can be any from 0 to 2·10⁹ inclusively. It is accepted for the base stations to have matching coordinates. If there are many solutions, print any of them.

题解

二分+贪心

二分功率，贪心验证能否成立……

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<iomanip>

#define REP(r,x,y) for(register int r=(x); r<(y); r++)
#define REPE(r,x,y) for(register int r=(x); r<=(y); r++)
#ifdef sahdsg
#define DBG(...) printf(__VA_ARGS__)
#else
#define DBG(...) (void)0
#endif

using namespace std;
typedef long long LL;
typedef pair<LL, LL> pll;
typedef pair<int, int> pii;

#define MAXN 200007
#define EPS 1e-3
int n;
int arr[MAXN];

char ch; int f;
inline void read(int &x) {
	x=0; f=1; do ch=getchar(); while(!isdigit(ch) && ch!='-');
	if(ch=='-') ch=getchar(),f=-1; while(isdigit(ch)) {x=x*10+ch-'0'; 
	ch=getchar();} x*=f;
}

inline int ub(int f, int t, double x) {
	while(f<t) {
		int m=(f+t)>>1;
		if(arr[m]<=x) f=m+1;
		else t=m;
	}
	return f;
}

inline int lb(int f, int t, double x) {
	while(f<t) {
		int m=(f+t)>>1;
		if(arr[m]<x) f=m+1;
		else t=m;
	}
	return f;
}

inline bool vali(double m) {
	int s=arr[0], x=0;
	REP(i,0,3) {
		x=ub(x, n, s+m);
		if(x>=n) {return 1;}
		s=arr[x];
	}
	DBG("#%d
", x);
	return false;
}

int main() {
	read(n);
	REP(i,0,n) {
		read(arr[i]);
	} sort(arr,arr+n);
	double l=0, r=1e9;
	while(r-l>EPS) {
		double m = (l+r)/2;
		if(vali(m)) {
			r=m;
		} else {
			l=m;
		}
	}
	printf("%.2f
", r/2);
	int s=arr[0], x=0;
	REP(i,0,3) {
		if(i) putchar(' ');
		int lx=ub(x, n, s+r)-1; printf("%f", (arr[lx]+arr[x])/2.0);
		x=lx+1;
		if(x>=n) {
			for(i++;i<3;i++) {
				if(i) putchar(' '); printf("0");
			}
			return 0;
		}
		s=arr[x];
	}
	
	return 0;
}