Can 1B exist any polylog solution?I guess the answer will no more than $$$O(n\log n)$$$ or $$$O(n)$$$ if all pairs of $$$a,b$$$ are distinct,but I have no idea how to find them in no more than $$$O(n\log^2n)$$$ :(

Maybe we can consider this round as Div1.5...2 hard 4 me.

Note that you only loop from 1 to 1e4,so it would goes wrong when n is large enough.

If the answer is No,it will run the loop completely,and your function f is $$$O(\log p)$$$,so the total is $$$t\times 10^4\times \log p=3\times 10^9$$$ and it goes tle.At least you should replace 1e4 with n*2.

BTW,I'm not sure whether it's right or not to use binary search.

You'll find that the task is just find whether there's an $$$y$$$ that $$$\frac{y(y+1)}{2}+x\equiv 0\pmod n$$$.

If n is odd,it's obvious that there's no influence to calculate with $$$y$$$ or $$$y\bmod n$$$,using $$$\frac{y(y+1)}{2}$$$ or $$$2\times(n-x)$$$ because 2 has its inverse.In that case since you find one,then it is correct and ends with f=0.So it's $$$O(n)$$$.

If n is even,it goes a bit complex.Set y as $$$kn+b$$$,where $$$1\le b\le n$$$,then recalculate the left and you'll get $$$\frac{k}{2}n+\frac{b(b+1)}{2}+x$$$,and in the first check both part were doubled so $$$+\frac n2$$$ has no influence.Then it has only 2 situations depending on $$$k\bmod 2$$$,so f could only be $$$0$$$ or $$$1$$$ and there's a result.Also it's $$$O(n)$$$

Nice E!And C,D are also interesting.

Anyway,what I'm dissatisfied with most is neither whether the round should be unrated,nor whether the provider is on purpose or not.

I'm really dissatisfied with is the attitude when a behavior break the rules.

Okay,it's hard to build up a contest,but now it really break the rules,and you tell me that because it's hard to prepare a round,you'll keep it rated?

That's not the point!The fact is that it really break the rule.

Since you have this rule and really used it several times in the past,why can you have this one exception,or you want to change the rule?

You can see most of them who solved F also solved A-D,or solved A-C very quickly.

If one cannot solve D and more,those people has very little influence to his final standing because even without F most of them is still ahead of him.

But for those ratings over 1800,especially for CM,they can always solve D.In that cases their standings will change a lot.And if you experienced CM times,you'll find even 100 standings different will cause a lot to your rating changes.

And obviously it's different from ICPC.

As an offline competition,you can't easily find the code,copy it and submit.Just like many problems that they have the same way to solve but a bit difference.Although you may know how to solve it as soon as you read the statements,you still must be very careful with the details in your code.

But as an online contest who cares?The solutions on the Internet can help you deal with these details and all you need to do is just to copy the code and submit.

and I also remembered there was a Div1 happened similar situations,and it was unrated.(my first participated Div1)And that was an AT's problem.

Why the two has different results?Because it's still unkown whether the problem provider did it on purpose or by mistake?(At least I don't know yet.Maybe I missed some important information.)

I wonder why many found F during the contest.I don't think it is so well-known a problem,and I may even didn't have a look at F during the contest(at least before I solve D).

Nice round.I enjoy it:)

Maybe E does not seem so difficult as F while they shared the same ratings,but that doesn't matter.Anyway,Happy New Year!