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。
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 xi coordinate. The village consists of n houses, the i-th house is located in the point with coordinates of xi.
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, ..., xn of integer numbers (1 ≤ xi ≤ 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
4 1 2 3 4
0.500000 1.500000 2.500000 3.500000
3 10 20 30
0 10.000000 20.000000 30.000000
5 10003 10004 10001 10002 1
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+10]; int n; int ans[4]; 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=1,r=n+1,mid=(l+r)/2; while (l<r) { if (loc[mid]<=y) l=mid+1; else r=mid; mid=(l+r)/2; } return mid; } int ok(int x) { int i,j; int z=1; for (i=1; i<=3; 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+1) return 1; } return 0; } int main() { while (scanf("%d",&n)!=EOF) { int i,j,k; for (i=1; i<=n; i++) { scanf("%d",&loc[i]); //cout<<loc[i]<<endl; } sort(loc+1,loc+n+1); //loc[n+1]=loc[n]; int r=(loc[n]-loc[1]); int l=0,mid; while (l<r) //二分覆盖圆的直径 { mid=(r+l)/2; if (ok(mid)) r=mid; else l=mid+1; } ok(l); //cout<<ans[1]<<" "<<ans[2]<<" "<<ans[3]<<endl; double radius=l*1.0/2.0; double radar1=(loc[ans[1]-1]+loc[1])*1.0/2.0; double radar2=(loc[ans[2]-1]+loc[ans[1]])/2.0;//前点加后点再除以2的方式得到圆心坐标 double radar3=(loc[ans[3]-1]+loc[ans[2]])/2.0; printf("%.6f ",radius); printf("%.6f ",radar1); printf("%.6f ",radar2); printf("%.6f ",radar3); } return 0; }