AGGRCOW - Aggressive cows

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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.

Solution: Binary Search

Well, the hint is already given in the SPOJ problem page. But it is confusing at the same time. Because you won't get the solution using Binary Search. Binary search will get you the optimal solution, but you need to still find a way to get all possible solutions.

Let Lim =1,000,000,000

To start thinking of a solution you need first choose the minimum difference between any 2 stable and then verify whether that minimum distance would work or not. Let suppose you have devised a function Fn(x), which tells you whether minimum distance x would is possible or not. Now what, now you will check this for all the numbers 0....Xi.....Lim.

0 is always possible since you can fill all the cows in one stable.

0, 1, 2 ...... Xi, Xi+1,......Lim

you will see that it is always possible to have a small number as a difference between stable, and as you increase the x you might reach a point where it is unstable. In other ways, you need to find Xi where Fn(Xi) is possible but Fn (Xi+1) is not. And X will be your answer. So to find X you can apply binary search on 0 to Lim and find X.

Now our job is to write Fn(X) and binary search which uses this Fn. 
Let me give you time to try this before I post my solution.

Find a pair of natural numbers who have the least energy among all pairs having sum of n

There is a natural number n. You have to find a pair of natural numbers x, y whose sum is n and also have the least energy among other pairs having the sum n. Then output the sum of digits of the pair.

Energy(x) = sum of all digits of x
Total Energy = Energy(x) + Energy(y)

1 <= n <= 10^9

For eg, 
n = 10000
A few pairs: 
5000 + 5000 -> Energy = 10
1000 + 9000 -> Energy = 10
9999 + 1 -> Energy = 37
2999 + 7001 -> Energy = 37

So possible answers are:
(5000, 5000), (1000, 9000) etc 
And Final answer is 10.

IT is not a SPOJ problem. I just came across this problem and wanted to share the tricky solution behind it.

Solution :

If n is of type 10^x then the answer is 10. Otherwise answer is the sum of digits of n.

The idea here is to break down the number into a pair containing digits less than that are present in n. if you break down into smaller digits then sum remains the same as the original number. example for 7= 1-6,2-5,3-4.

for a number like 100, 1000.... digit 1 can’t be broken down into further pairs, so we try to make 10 as the sum of digit so that the sum becomes n. like for 10=5-5,2-8,3-7 100=20-80,40-60

for other numbers, like 123 it can be broken into 100-23, 120-3, 111-12... all will give you sum 6. which is the sum of digits of the original number.

if you try to break down into further pairs like 80-43, 52-71, you will see that the digit sum increases as you have broken down to a number containing digits which are higher than those are present in n. like 8 4,5,7 are greater than 3.