d你这就是玄学贪心呗，从大往小拿边，因为大的能拿的边小的也能拿，至于会不会出现NO情况你也不正面，别downvote啊，downvote就是急了

不好意思长官，刚才声音太大了，其实也不用那么“注意到”

求求中国人别出题了，题解都是 “注意到”，“很简单”

In D, why the best ans is the time of root is 1? how about other condition? if can not, should we use change root dp to fix it?

For problem C,Can someone provide me the prove of the (m-c)!/((m-c)/2)! is the case number. It seems very interestring. Thanks

这D2的代码和题解真的弄的是一道题？后面的公式全是错的，真的，不想写可以不写的

写题解不写完整是吧，证明留给读者，我真艹了你*了

老子真草了，题出这么难干什么啊？

div2 F's solution is so cool!

So, CartesianTreeForces!

Could someone tell me how to solve F...Thanks!

For F:

Can we just calculate every one's contribution independently,let's denote we make I be the winner and I have defeated the S(include I) and we want to defeat j , the transition is dp[S|(1<<j)] = mul(dp[S|(1<<j)],f[S][j]) ,f[S][j] means that the probability defeat j with at least one member in S.

How can we hack the thought.... I code it and fail on the test2... Can someone hack me,please,I almost be mad....

It is really a ****ing enjoyment to read your code for F(div2)...orz %%%%

Will the rating roll back QAQ

Why so many downvotes.... I think the contest is very nice

I think I have a better solution for F in implement,if the n is 5000 , we can use HashMap or other "almost" O(1) datastruct to replace map.Let's get down to our business,we just naive denote dp[i][j] is we finish i task and we left j bullets , we cost the minmum bullets , and normal transtation which should consider if we finish the i in advance or R[i]!=L[i+1] we can refresh the dp[i+1][k] the Code: https://codeforces.me/contest/1430/submission/95434956

Because I have seen one named "Forced online Query" can be solved by offline method QAQ

I just think the best name for problem F is "The real Forced online Query"

When I read the solution for F,I just want to laught at myself....Nice problems!Thanks.

So,I think I have a more clear solution for D.For every [l,r] which is 0 ,we want to insert as few as possible numbers to break them,we have an idea that find all the [l,r] which equals 0 and select as few as possible points to "touch" all the intervals(which I means that makes every interval has at least one point) and we find it is O(n^2) but the greedy model is classical,we sort the intervals and select as right  point as possible and we find that for [l1,r] [l2,r](l2>l1) if we can touch [l2,r] we can touch [l1,r] so,we just store the [l2,r]=0 which the l2 is the biggest.

Sorry for my poor English,and it may sound silly...But I hope it will be helpful for you bros

catch NWU's juju

wow,E is so beautiful ,although the EDU is often difficult and full of naughty and lovely hackers,the problems can always make me marvel at them! Thanks for neal's solution and the nice contest!

Hey,bros,why down vote . I just express mine and many guys'

Nice solution,make full use of the feature of the problem and the prefix add and minus make us avoid discussing the conditions and the important point is that the "restriant" that the min possible value will no less than 0 and what we want to count is just 0 . Just because of the feature we can use segtree maintain it.I just hope that I will never compete with you.... :D

When can I be so smart as you errorgorn orz....

Why Unrated！！！We can just extend the length of the contest

Firstly,really thank you for your forgiveness on my rudeness and your deeper explaination. I somehow understand what you have said,and I will have a try.Thanks for your explaintion and patientenss again!(sorry for my poor English....)

Good tutorial,but the div2E's solution is really vague,"why We will always finish the game at stage n or n−1"? The explaination can never be understood,let's think we have n = 5,1->2->3->4->5->4->3->2->1 must be some data set's optimal solution.I am silly,and you guys can downvote me for my rudeness, but I really hope the offical can explain this clearly!!!

if the edge <= 10 I will solve it...

E is a really nice problem ! Thanks,AlFlen and 74TrAkToR!

Sorry,You have been red

Yes.... the fact that everyone can be grandmaster is never true(I think I can never be..).But I think hard work is always important.I think your frank can help you become red one day.So,I upvote

Can some cool guy explain the fact that when anyone ask "Is it rated",he must get many downvotes

The worst people are those who have not very high rating,looking down upon those who have lower rating,but diligent guys

Sorry Sir,I am too excited with int128's appearance in codeforce ,I think I should apologize for every pyer and javaer~ Sorry bros!

emmmm,Sorry

So,bros,Can we solve F in this way.We find bccs in the graph and check whether there are a bcc,whose size is not less than sq,if not we sort the nodes as their degree,and add as many nodes into the independent set as possible. And I have some implenment problem. If we know the nodes in a simple circle,how can I print them in the order.Can someone answer me,Thanks bros!

Can someone explain the div2.E's the way to maintain the number of the red and blue nodes,I can not understand the code's defining of l,r...

Let us solve D by dp!!! you may say are you crazy?Be patient,my friend we can naively denote the status dp[i][e][a][b][c][d] is when we are at i position and the end is "e"(0,1,2,3) element we has a zeros ,b ones ,c twos ,d threes ,so the transform is obvious so, we have many status....TLE? NO,many status can just be neglected let us denote x,y is the element we want to put ,and denote nd:=min(cnt[x],cnt[y]) nd+1,nd is the number we can put ,other condition in fact will finally transfer into the status my code: Your text to link here...66490372

So,for problem D I have the proving process, and the Solution ignore it,the code ID:66302024 Welcome to Hack~

I just hope there are 2:15 minutes

emmmm.... I misunderstand the problem,sorry

I can hack you with the following data:

the origin string is "abd",3 operations + 1 a + 1 d + 2 b the answer should be YES YES NO but yours is YES YES YES

I think the transition has problem:

when string is : "abdabc" , and then the s[0] is "ad", the s[1] is"b",s[2] is empty,the dp[2][1][0] should be 4 but the program show the dp[2][1][0] is 2,and the problem is caused by the wrong transition:dp[1][1][0] can not update the dp[2][1][0] by just using the helper arry

But I my opinion , the statue may be dp[0][i] represent that "The i-th element is in increasing sequence and the max possible element in the decreasing sequence",and dp[1][i] represent that "The i-th element is in decreasing sequence and the min possible element in the increasing sequence" May be I am Wrong ,But I hope you can correct my mistake , Thank you