二分 例题3

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 4932

Description

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.
代码：
为什么要加后面那一段再判一下呢？删掉就错了呢？为什么呢？我也不知道啊，是吧？

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;

const double inf=1e9+7;

int n;
double a[56];

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

int main()
{
int T;
int i,j;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%lf",&a[i]);
sort(a+1,a+n+1);
double lb=0,ub=inf;
for(i=1;i<=100;i++)
{
double mid=(lb+ub)/2.0;
if(C(mid))
lb=mid;
else
ub=mid;
}
for(i=1;i<n;i++)
if(C(a[i+1]-a[i]) && (a[i+1]-a[i])>lb)
lb=a[i+1]-a[i];
printf("%.3lf\n",lb);
}
return 0;
}

View Code