Thank you for the information, the contest was rescheduled.

Uploaded the editorial.

F looks very hard but after a single observation it looks much easier.

1500 is partial point for the tree case.

10 10
1 3 1 3 8 1 3 1 3 8

Your answer is 32.

The right answer is 24:

* * * * * * * * * *  // all places unpaid
1 * 1 3 * 1 * 1 3 *  // pay for 6 places, total cost = 10
* Y * Y Y * Y * Y Y  // shift, total cost = 20
1 Y 1 Y Y 1 Y 1 Y Y  // pay for 4 places, total cost = 24

if((a[0]%2==0)||(a[0]%2==0)||(a[0]%2==0))

The above line from your code for A doesn't look right, does it?

And why would your B and C solutions work? The logic is wrong in both.

It is supposed to be 256MB (see stack limit section), do you think it works properly?

I didn't understand the editorial of B either...

Can anyone use an example to explain? Thanks a lot.

Here we consider a case where we cast exactly K magics. Of course it is possible to just captuer a slime of color 3, and this case is covered by K = 0.

Consider that you shift exactly 2 times. For slime i,

1. If you want to capture it directly, capture it after you shift 2 times.

2. If you want to get it from slime(i-1), capture slime(i-1) after you shift 1 times.

3. If you want to get it from slime(i-2), capture slime(i-2) before any shift.

If you fix the number of times you make a shift, the cost of shift is fixed. Therefore you only need to consider which kind of slime should you catch to get a slime of color i.

You are declaring A B C as integers and then multiplying them before casting them to unsigned long longs. ex: 999999999 999999999 999999999, your code prints out 808348673 while the answer should be 999999998000000001.

Try this: http://ideone.com/oDtVeg.

Try test

8 2 1 1 2 3 3 3 3 3

Answer 1