AGGRCOW - Aggressive cows


Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).

His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ wants to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

Input

t – the number of test cases, then t test cases follows.
* Line 1: Two space-separated integers: N and C
* Lines 2..N+1: Line i+1 contains an integer stall location, xi

Output

For each test case output one integer: the largest minimum distance.

Example

Input:

1
5 3
1
2
8
4
9

Output:

3

Output details:

FJ can put his 3 cows in the stalls at positions 1, 4 and 8,
resulting in a minimum distance of 3.


hide comments
hellb0y_suru: 2019-12-17 20:39:46

AC in one go !!!
good question on BinarySearch !!!
must do !!

saurav_03: 2019-12-08 10:10:49

interesting problem

saurav_23: 2019-10-26 08:52:01

AC in one go! Good question but sample test case could've been better.

gadela_kesav4: 2019-10-17 18:34:03

Test cases are not up to mark showing correct answer even if coded wrong

dhairya_2000: 2019-09-29 13:22:29

Finally One Correct in One Go!!

abitnav: 2019-09-28 15:29:56

Last edit: 2019-09-28 15:34:15
jpgmoreira: 2019-09-12 00:27:11

Really good question about binary search.

kb2201: 2019-09-02 17:08:23

great ques but poorly worded

arpitjp: 2019-08-25 12:10:34

My code working on my computer but showing wrong answer in SPOJ

#include<bits/stdc++.h>
using namespace std;
int barn, cow, temp;
vector<int> pos;

bool place_them(int mid)
{
int count=1, start=pos[0];
for(int i=0; i<barn; i++)
{
if(pos[i]-start >= mid)
{
count++;
start = pos[i];
}
if(count==cow) return 1;
}
return 0;
}

int binary(int lo, int hi)
{
int mid;
while(hi-lo>1)
{
mid = lo + (hi-lo)/2;
if(place_them(mid)) lo = mid;
else hi = mid;
}
return mid;
}

int main()
{
int t;
cin >> t;
while(t--)
{
cin >> barn >> cow;
pos.clear();

for(int i=0; i<barn; i++)
{
cin >> temp;
pos.push_back(temp);
}
sort(pos.begin(), pos.end());
int lo, hi;
lo = 0;
hi = pos[barn-1] - pos[0] + 1;
temp = binary(lo, hi);
cout << temp << endl;
}
return 0;
}

Last edit: 2019-08-25 12:11:09
abeshek_2000: 2019-08-17 10:58:01

great question on binary search!!


Added by:Roman Sol
Date:2005-02-16
Time limit:2s
Source limit:10000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All
Resource:USACO February 2005 Gold Division