Who cares

Yes

It seems that's right.

I think it's for admins to give announcements in case website crashes or something.

You have replied to a comment which got a lot of downvotes; such comments are hidden as well as replies to it

definitely yes.

Radewoosh vs V--o_o--V vs scott_wu

i think mnbvmar going to be the first in rating

No, this contest isn't rated for users from rural parts of Yugoslavia.

lol

This is codeforces, wait for some days then you can see the list of the contests.

so please be with codeforces.

Is this the reason why so many submissions on D or is it a very standard problem????

Seems like n = 1.

My code fails on this case, nobody hacked me :(

RIP my rating.

If you can count number of numbers with remainder i, it's easy to calculate number of x*x % m = i.

We just have to store the values of (i*i)%m for i=1 to m, the values repeat thereafter.So maintain an array which will store number of values for a particular mod value(0 to m-1). a[(i*i)%m]+=(n/m) for i=1 to m and then compute for remaining values(i=(n/m)*m+1 to n). Then,loop through all values of mod such that mod value k can pair with mod value (m-k).

How to solve D ?

If we knew the number of elements in range [1², 2², ..., n²] that are divisible by m with remainder i (where, 0 <  = i < m), we can esily count the result, that is, the number of pairs of squares elements that are their sum is divisible by m (without remainder).

To know that, we should know:

1) how to find the number of elements in range [1, 2, ..., n] that are divisible by m with remainder i (where, 0 <  = i < m).

2) how to transform from [1, 2, ..., n] to [1², 2², ..., n²].

This is my submission for more details.

I got RTE on 5 because didn't use long long for some multiplications, maybe that's your case?

Big strings' sizes i think. My solution failed because of overflow. Used long long and it passed.

I think anti hash.. I changed my hash function and it passed for me..

Probably some random test case, but looks like I can only pass by switching to using binary search / ternary search instead of analytic solutions.

For D: count the number of leaves in every subtree, then sort.

How to solve D?

((i*i) + (j*j))%k can be written as ((i%k*i%k)%k + (j%k*j%k)%k)%k. Keep count of number of i's such that (i%k) = j for all j from 0 to (k-1). Similarly using this count array, construct (i*i)%k mapping to j, let this be arr. Now for every i,j such that 0<=i<k and 0<=j<k if (i*i + j*j)%k == 0 add the value of arr[i]*arr[j].

Instead of thinking of the coordinates as numbers from 1 to n, think of it as several consecutive periods of numbers from 1 to m.

Traverse an integer i from 1 to m. With each value, calculate two value:

• cnt: number of periods that contains i within range [1, n].
• r: the remainder of i² when divided by m.

Make an array RemCnt[], and with each i, add cnt to RemCnt[r].

After this, we can safely do a O(m²) traverse to find out the number of pairs (i, j) such as i + j is divisible by m.

Hint please

I solve it very nicely.

Very good problem.

True that. But I felt it was harder than both C and D.

What's output for n=1?

Probably you forgot to print an empty line, remember that the array of parents must be printed in a different line.

Haha, May the force be with us all :P

No reason to be afraid of system tests if you survived hacking phase, I didn't have that luck ;_;

I'm using an insanely weird modular number, yet the "single" property still makes me worried as hell, especially seeing Swistakk got hacked...

Can you explain your solution for E?

Probably the query you are accidentally printing is the value of the player itself and not the index. I got a WA on test case 7 for that one.

I think for optimal solution you have to take special pairs of heroes first. I also got WA on test 7..... I fix this and pretest passed.

Simply you were getting the strategy wrong.

If you have a turn and not being obliged by anything, you can choose one pair of special heroes, pick the stronger one, therefore forcing your opponent to pick the weak one, and give you back your turn again. It's a clear win-win, so whenever you have a chance to freely choose heroes, repeat that process until there's no special heroes' pairs left unchosen.

After that being done, we return to the normal, maximum-prioritized strategy.

We color leaves only.

"all the lights" so the nodes with no lights are ignored, every junction is concerned only with the "lights" in its subtree.

I think hack for D is n = 1

Печально, если так :/

Because it is corner case and it should be quite obvious to check before submitting solution. So if you didn't that — it's your fcking problem.

Depending on how you implemented it, n=1 is not even a corner case; it can be handled by the general case. If you consider the sole node in n=1 to be a leaf and begin processing nodes at the leaves, you shouldn't run into an error with n=1, and you don't need a special case for it either.

The only thing you should do is to stop complaining and improve yourself, and you will find out that pB is trivial enough.

rsFalse has a point here. B, C, D felt of the same difficulty, which isn't good already. For div. 1 guys they all felt easy, and for div. 2 guys they were all medium. Making B easier and D harder would've been appreciated by more people as then B would've been solvable by more div. 2 and div. 3 people, while D would've been more interesting for div. 1 participants. Problems of the same difficulty really create issues, but this is difficult to avoid when preparing a contest.

For a given n and m you have to count how many i, j are there where (i^2 + j^2) = 0 (mod m). You can easily see that i^2 mod m and j^2 mod m ( i, j from 1,..., n) forms a cycle.So, ans will be equal cnt[i^2 mod m] * cnt[m — ( i^2 mod m)]. Check my submission 46215431

Maybe they are just taking care of new problems/competitions before posting them.

? Time Limit is 3sec, n is 1e6 and intended solution is n.

Did you mean H?

You answer is incorrect since you have 64.

Notice the first numbers from 2nd line and so on indicate the number of stops in that line and it is not a stop itself.