HDU4932——二分——Miaomiao's Geometry

There are N point on X-axis . Miaomiao would like to cover them ALL by using segments with same length.

There are 2 limits:

1.A point is convered if there is a segments T , the point is the left end or the right end of T.
2.The length of the intersection of any two segments equals zero.

For example , point 2 is convered by [2 , 4] and not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3 , 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of intersection doesn‘t equals zero), [1 , 3] and [3 , 4] are not(not the same length).

Miaomiao wants to maximum the length of segements , please tell her the maximum length of segments.

For your information , the point can‘t coincidently at the same position.

Input

There are several test cases.
There is a number T ( T <= 50 ) on the first line which shows the number of test cases.
For each test cases , there is a number N ( 3 <= N <= 50 ) on the first line.
On the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the position of each point.

Output

For each test cases , output a real number shows the answser. Please output three digit after the decimal point.

Sample Input

3
3
1 2 3
3
1 2 4
4
1 9 100 10

Sample Output

1.000
2.000
8.000

Hint

For the first sample , a legal answer is [1,2] [2,3] so the length is 1.
For the second sample , a legal answer is [-1,1] [2,4] so the answer is 2.
For the thired sample , a legal answer is [-7,1] , [1,9] , [10,18] , [100,108] so the answer is 8.

/*
大意：在n个点之间放线段，问最大多少
去重
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n, T;
const  double inf= 1e9 + 100;
const int MAX = 55;
double a[MAX];

bool check(double x)
{
    double cur = a[1];
    for(int i = 2; i <= n ; i++){
        if(cur > a[i]) return false;
        else if(cur == a[i]) continue;
        if (a[i] - cur >= x){
            cur = a[i];
        }
        else cur = a[i] + x;
        }
    return true;
}

int main()
{
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(int i = 1; i <= n ;i++)
            scanf("%lf", &a[i]);
        sort(a + 1, a + n + 1);
        double l =  0, r = inf;
        for(int i = 1; i <= 100; i++){
            double mid = (l + r) / 2 ;
            if(check(mid)){
                l = mid ;
            }
                else  r = mid ;
        }
        for(int i = 1; i < n; i++){
            if(check(a[i+1] - a[i]) && (a[i+1] - a[i] > l))
                l = a[i+1] - a[i];
        }
        printf("%.3f\n", l);
    }
    return 0;
}

HDU4932——二分——Miaomiao's Geometry