CodeForces 51C 二分搜索

校队选拔神马的事情就不说了，哥们反正是要崛起的人了！

感谢何骐的提醒。

校队选拔的时候又被二分给坑了，所以还想做几道二分搜索的题目来练练手。

C - Three Base Stations

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 51C

Description

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.

Sample Input

Input

4
1 2 3 4

Output

0.500000
1.500000 2.500000 3.500000

Input

3
10 20 30

Output

0
10.000000 20.000000 30.000000

Input

5
10003 10004 10001 10002 1

Output

0.500000
1.000000 10001.500000 10003.500000

初看这个题目的时候，我以为又是切蛋糕那种题目，二分一个浮点数，（事实上根据网上博客贴的代码，有人这样做也可以过）。但是后来接触到一种思路，事实上点坐标都是整数，所以站点的覆盖范围，即覆盖圆的直径，必定是一个整数，故，只要二分覆盖圆的直径，即可。

关于最后还要输出每个站点的具体坐标，一个好的方法是在二分时用个数组记录每建立一个覆盖点后，覆盖到的点的下一点。这样，求坐标的时候，还是在代码里面说吧

比较坑的是第一组数据，我为这个纠结了好久，明显的在第一组数据中，两个站点就可以覆盖4个点，第三个站点随便怎么放，但我以为一定要按样例，改来改去，都没法统一

结果发现根本就不用管这个破第一组样例，也过了。可能像这种数据，随便站点怎么放，只要能覆盖就过了吧。

#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 200000
#include <algorithm>
using namespace std;
int loc[maxn+];
int n;
int ans[];
int fnext(int y) //用来找到当前覆盖圆覆盖后的最近一个覆盖点。
{
    int l=,r=n+,mid=(l+r)/;
    while (l<r)
    {

        if (loc[mid]<=y)
            l=mid+;
        else
            r=mid;
        mid=(l+r)/;
    }
    return mid;
}
int ok(int x)
{
    int i,j;
    int z=;
    for (i=; i<=; i++)
    {

        z=fnext(loc[z]+x);
        ans[i]=z; //用这个数组来记录覆盖圆的下一个点。
        //cout<<x<<" "<<z<<endl;
        if (z>=n+) return ;
    }

    return ;
}
int main()
{

    while (scanf("%d",&n)!=EOF)
    {
        int i,j,k;
        for (i=; i<=n; i++)
        {
            scanf("%d",&loc[i]);
            //cout<<loc[i]<<endl;
        }
        sort(loc+,loc+n+);
        //loc[n+1]=loc[n];
        int r=(loc[n]-loc[]);
        int l=,mid;
        while (l<r) //二分覆盖圆的直径
        {
            mid=(r+l)/;
            if (ok(mid)) r=mid;
            else l=mid+;
        }
        ok(l);
        //cout<<ans[1]<<" "<<ans[2]<<" "<<ans[3]<<endl;
        double radius=l*1.0/2.0;
        double radar1=(loc[ans[]-]+loc[])*1.0/2.0;
        double radar2=(loc[ans[]-]+loc[ans[]])/2.0;//前点加后点再除以2的方式得到圆心坐标
        double radar3=(loc[ans[]-]+loc[ans[]])/2.0;
        printf("%.6f\n",radius);
        printf("%.6f ",radar1);
        printf("%.6f ",radar2);
        printf("%.6f\n",radar3);
    }
    return ;
}