Mad cow

The time limit :1000 ms  |  Memory limit :65535 KB
difficulty :4
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 ?
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
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
Sample output

#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
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[n-1]-a[0],mid=0;
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 @
max=mid-1;// can't let go , It has to be reduced mid Value
return min-1;// because @ It's about min+1, So here minus one

int main()
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
return 0;

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