C - Three Base Stations

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

     

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

41 2 3 4

Output

0.5000001.500000 2.500000 3.500000

Input

310 20 30

Output

010.000000 20.000000 30.000000

Input

510003 10004 10001 10002 1

Output

0.5000001.000000 10001.500000 10003.500000

本题有两种思路：一种，枚举三段，找最小，然后优化，第二种，使用二分距离判断是否符合条件。本题有个隐藏条件，那就最终的距离只可能是前面一个整数，后面跟个0.5或0.

#include
    
     #include
     
      #include
      
       #include
       
        using namespace std;const int mm=2e5+9;const int oo=1e9;int dis[mm];int n;double a,b,c,len=oo;void ok(int x,int y){ if(y<0)y=0;  int z;  z=dis[x]-dis[0];  z=max(dis[y]-dis[min(x+1,n-1)],z);  z=max(dis[n-1]-dis[min(y+1,n-1)],z);  if(z
        
         dis[j]-dis[i+1])        ++j;      ok(i,j);ok(i,j-1);    }    len/=2.0;a/=2.0;b/=2.0;c/=2.0;   printf("%.6lf\n%.6lf %.6lf %.6lf\n",len,a,b,c);}
        
       
      
     
    

二分代码

#include
    
     #include
     
      #include
      
       #include
       
        using namespace std;const int mm=2e5+9;const double oo=1e39;const double ex=1e-7;double locat[mm];int n,ans[mm];int blook(int x)///返回比x大的第一个数的下标{  int l=0,r=n,mid=(l+r)/2;  while(l
        
         x)r=mid;    else l=mid+1;    mid=(l+r)/2;  }  return mid;}bool ok(int x){ int z=0;  for(int i=0;i<3;i++)  {    z=blook(locat[z]+x);    ans[i]=z;    if(z==n)return 1;  }  return 0;}int main(){  while(cin>>n)  {    for(int i=0;i
         
          >locat[i]; sort(locat,locat+n); int r=2e9+9,l=0,mid; while(l