Mad cow
 describe

Farmer John Built a long corral , It includes N (2 <= N <= 100,000) Compartments , The compartments are numbered in turn x1,...,xN (0 <= xi <= 1,000,000,000).
however ,John Of C (2 <= C <= N) The cows don't like this layout , And a few cows in a cubicle , They're going to fight . In order not to let cattle hurt each other .John Decide to allocate compartments to the cattle , Make the minimum distance between any two cows as large as possible , that , What is the maximum and minimum distance ？
 Input

Multiple sets of test data , With EOF end .
first line ： Two integers separated by spaces N and C
The second line —— The first N+1 That's ok ： It is pointed out that xi The location of  Output
 Each group of integers outputs one test data , Satisfy the maximum and minimum of the meaning of the problem , Pay attention to the new line .
 The sample input

5 3 1 2 8 4 9
 Sample output

3
#include <iostream>
#include <algorithm>
using namespace std;
long long a[1000000];
int n,c;
bool Judge(int mid)
{
int i,sum=1,temp=a[0];// At least one cow
for(int i=1;i<n;i++)
{
if(a[i]temp>=mid)// The distance from the first room number is once
{
sum++;
temp=a[i];// to update temp Value ,temp To a[i] You can put down a cow in between . temp=a[i], Ask for the next a[i] To temp No, you can put down a cow
if(sum>=c)// If the total sum>c, You can put it down c Head ox
return true;
}
}
return false;
}
int Binary_search()// The maximum of the minimum distance of binary search
{
int min=0,max=a[n1]a[0],mid=0;
while(min<=max)
{
mid=(max+min)/2;// Take the value between the minimum distance and the maximum distance first
if(Judge(mid))// If the current distance can be lowered c Head ox
min=mid+1;// Try to increase the minimum distance @
else
max=mid1;// can't let go , It has to be reduced mid Value
}
return min1;// because @ It's about min+1, So here minus one
}
int main()
{
while(cin>>n>>c)
{
for(int i=0;i<n;i++)
cin>>a[i];// The position of the smallest number
sort(a,a+n);// Sort these minimum numbers in ascending order
cout<<Binary_search()<<endl;
}
return 0;
}
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