yeah it will be one atrocity of a contest. this guy has never made a contest before so we can expect the worst...they are just making fun of us at this point

I think 4, but it is not certain.

I also hope)

Separate

There is actually a link in the post.

But why do you incur receiving negative vote by yourself? Your tone is the reason why there are a lot of negative vote on your comment. The same will be true of other comments. As I checked, 47 of your 60 comments received negative vote, 11 received zero sum of positive and negative vote, and only two received positive vote. And few people will think well of your comments though you may be going to post more comments in that tone.

Good luck!

I am really sorry, that your preserved color is gonna fade away :|

Oh,404Error......

Hacks are part of codeforces contests

There has been a spate of weak pretests in recent contests.

Hmm, no, have you been too obsessed with CodeCrafts?

AFAIK two most recent contests are fine.

jinlifu's round is a pretty good round ruined only by server issues.

And about my contest, please research closer (all following data is calculated during contest time):

No, we would say if there was an interactive problem

really no one cares

It is better to refrain from containing "Is it rated?" in the comment.

It's always a bad time for someone, quit whining.

I had that feeling too in my first Div. 1 round, luckly things ended up great and I solved 2 problems earning +76 :D

Good luck

I'm in the same boat (@1904). I'm both nervous and excited! Good Luck!

The round will be rated for both divisions.

Are you gay?

Best of luck to you also

How can sleeping time be bad ?

try: 2 100 1. should give you 20.

Atmost 1 operation can be performed by the farmer. Trying to perform more than one operation results in failure in this test.

There's no matter in how you place the brackets

Does someone have to post this every time?

What the hell "@yourfavoriteexgf" is doing in that image??

The boundaries of the range described

I saw a person with +3:-2.

You can always go to the contest->status and use status filter to see all successful hacks in the contest.

Your solution might have a bug.

tle in 11 worst case takes around 3 seconds tried every optimization :(

Though I didn't participated but I think dividing prefix xors for even and odd indexes should work

This might help:

Let's say you have numbers A and B. C = A XOR B Then C XOR B will be A. You can use this in ranges too

Let's consider segment (l, r] where r-l is odd, then
p[r] xor p[(l+r)/2] = p[(l + r)/2] xor p[l]
p[r] = p[l]
Now it's simple thing to calc.

Maintain prefix XOR. Partition the indices with two of them in the same set if their prefix XORs are same. Maintain separate count for odd and even indices in each of the partitions. Answer will be the sum over all partitions of (odd choose 2) + (even choose 2).

First, note that the problem can be reduced to count number of subarrays given xor by noticing that

a[i] ^ a[i+1] == a[i+2] ^ a[i+3] equals to a[i]^a[i+1]^a[i+2]^a[i+3] with zero as a result.

Hence, given xor will be 0.

Since the problem limits to even-size array only, we need to modify the array a bit. We can try to merge every index i and i+1 (i.e. a[1]..a[n] becomes {a[1]^a[2], a[3]^a[4], ... } Then the problem simply becomes count number of subarrays given xor.

But, another caveat is that we only handle even indices. We also need to create an array consisting of {a[2]^a[3], a[4]^a[5], ... } to handle odd indices.

P.S. You can simply google how to count number of subarray given xor

Hint : Answer belongs to {1, 2, Impossible}

Assuming my solution works:

You can split the problem into two cases, a palindrome of odd length and even length. For both cases, you can notice that the answer will either be 1 or two. That's a major hint.

For odd cases: Checking for the "Impossible" case isn't too difficult so I won't explain that. The answer will always be 2 for this case otherwise (we can cut a prefix and a suffix of the string a "swap").

For even cases: If both halves aren't palindromes, the answer is 1. Else, we can check if we can cut some prefix and append it to the back to form a palindrome that is different from the original one (if yes, the answer is 1). Otherwise, the answer is 2 by the same reasoning as the odd case.

Div2D: Solutions fall into 3 cases: {1,2,impossible}.

Let's think of a palindrome as consisting of "mirrored" part and a possible "center" character. For example, "abcxcba" has "x" center and "abc" mirrored part.

Impossible case: input where the mirrored part consists of only 1 character. For example, "qq" or "qqqxqqq". If the mirrored part consists of more than 1 character, for example, "abcxcba", then we can always solve with 2 cuts (cut the outermost 2 characters from both sides in the example).

So if the answer is not impossible we just need to check if a 1-cut solution is possible. If it's not, the answer is 2. Since input length is only 5000, we can try all possible places for 1 cut, check that we get a palindrome, and that the palindrome is different than the original palindrome.

Can you explain D?

You should not take into consideration events that appeared before l. In other words, the initial s on the interval should be 0, not the last speed that was set before l. Such a fix changed my score from "Wrong Answer on test 9" to "Pretests Passed".

https://en.wikipedia.org/wiki/Cayley%27s_formula #generalizations

Then you just enumerate the number of points between A and B

Knowing how to google properly gives 1750 in div1! How motivational

Use cayley's and basic combinatorics to find the number of trees in which we have k edges in simple path from a to b. (a, b doesn't actually matter)

let's renumerate vertices a = n — 1, b = n. Than think about Prufer Code. You'll have path between n — 1 to n with number of edges >= e if Prufer code ends with (e — 1) sequence of distinct numbers that are less than n — 1. Than you can find formulas for dp[e] (it's just what is written above) to make number of edges == e just subtract from dp[e] , dp[e + 1] and dp[e + 2] ... Implementation for details 50017868

Sad thing is I looked at that page but skipped past the formula. Sad face!

how u solved div1 b ?

i dont know if this is helpful, but testing your code against: 2 1 100. gives me 100 instead of 20.probably you're missing the fact that your meant to perform the increase & decrease at most once.

forn(i,0,n)
    {
      cin>>a[i];
...

 forn(i,1,n)
       {
...

second for is from 1

for div1d , have u used cayles theorm ?

Поддерживаю тупо со смыслом.

#define read(a,n) upto(i,0,n-1,1) cin>>a[i];
...
for (int i = n-1;i>=1;i--)
...

for to 1 instead of 0

What will you do when all numbers are odd?

What about this one?

5

1 3 5 7 64

Optimal X = 8, operated on 64

4
7 5 8 3 

answer is 22, your program is giving 23

Next time you want to put some code in comments, I suggest you using pastebin, ideone, link to your submission or even the thing called "spoiler", because it's not very nice to put your long code in the comments section)