maybe delete this contest? if you agree upvote me

We increase the number of problems by adding some new problems or subtasks of existing problems to smooth some gaps in difficulty.

Lolololollolollllll, who asked you to divulge the story of your life??

Never give up. One day we 'll change our colour.

Thanks for the billion dollar information. You saved the world.

I don't think it becomes true someday :(

Div.3 are much less useful for most participants so I gain less contribution than this tiny anime girl.

Well guess div3 always gets a judging queue that ruins the contest.

Yes, I will.

As you see, I have AKed the contest. It means that I will AK IOI soon.

Well, I guess that Technocup Elimination Round 1 announcement would be a better place for such question.

Yes,It is ! :)

you downvoted me.... but I still will show my strength in the contest!!! (angery smiley)

problem A

exactly!

NO , I'm just another account of Alex_Wei

And AIex_Wei is a FAKE one , he is chenxiaoyan

Какая разница, какая из задач содержит в себе подзадачу, если это сглаживает переходы по сложности между ними?

How to solve F? I tried solving it using trie greedily but later figured out that greedy won't work.

Yeah an application of SOS(sum over subsets) dp. Link to the blog is :

Link

Drake no: Screenshot

Drake yes: Photo of screen

this legit happened to me today

nice :)))))))))

there are 26 lowercase characters I used bitwise OR segment tree then count the number of bits for the L,R range

Your n is not initialized anywhere in the code.

int BIT[100002][26], a[1000],t,n;string s;
void update(int x, char ch)
{
    int ind=int(ch-'a');
      for(; x <= 100001; x += x&-x)
        BIT[x][ind] ++;//,cout<<ind<<" "<<BIT[x][ind]<<" hh\n";
}
void delet(int x,char ch)
{
      int ind=int(ch-'a');
      for(; x <= n; x += x&-x)
        BIT[x][ind] --;
}

You can solve it easer. Just take a 26 sets of $$$i$$$th character positions ($$$i$$$th set contains positions of $$$i$$$th alphabet char in string). So, when you've got a update query, just delete $$$pos$$$ from $$$lastc$$$th set, and add $$$pos$$$ to $$$c$$$th set. When you've got a count request, for each of 26 charactes find any index in set, where $$$i\geq l$$$ and $$$i\leq r$$$, if it exists, just add one to answer.

I tried to solved D by simply map and string. I thought it will got TLE but I got AC. Here is my code. Can anyone suggest what is complexity for my code?

There is one way from $$$(1,1)$$$ to $$$(2,n)$$$ in any dataset, for which answer is "YES", only .

I have used the simulation strategy to solve this problem:

start from column 1 and row 1 :

if(pipe of type 1 or 2): you have to just move to next column with the same row else if(pipe of type 3,4,5 or 6): ... check the other row with the same column should have pipes(3,4,5 or 6), if(check = true): just change the value of row i.e from row=1 to row 2 or from row=2 to row 1 else if(check = false): you should print no as final answer

at last if row == 1: you should print 'NO' else The final result remains as it is what you get from the previous steps

You can understand better with my code: q=int(input())

for _ in range(q):

n=int(input())

s1=input() #column 1
s2=input() #column 2

r=1  #value of row
ch=1 #variable to check the results during simulation


for i in range(n):

    if(r==1):
        if(s1[i]=='1' or s1[i]=='2'):
            ch=ch
            r=1
        else:
            r=2
            if(s2[i]=='1' or s2[i]=='2'):
                ch=0

    else:
        if(s2[i]=='1' or s2[i]=='2'):
            ch=ch
            r=2
        else:
            r=1
            if(s1[i]=='1' or s1[i]=='2'):
                ch=0



if(ch==0 or r==1):
    print("NO")
else:
    print("YES")

I solved C using recursion, but was able to submit after 20 min of contest(my bad was not able to figure out one error)

Usually during contest, i coded with comments so i don't get lost for what i need to do.You can check my submission here: https://codeforces.me/contest/1234/submission/61665005

lol same wasted 2 hours over this. All solution on B2 should be rejudged everyone knows unordered_map is faster than map . It costed me almost 1000 rank due to this shitty issue.

me too.o(╥﹏╥)o .So I didn't solve B2 in the contest .And I want to know why the unordered_map is less efficient than map this time.

Even I wasted almost 30 min on this to recall that the worst case complexity of unordered map or hashmap is O(n) i.e. linear time...

Kudos to B2 problem to refresh my basics. But unfortunately I'll be afraid of using it in future

Do not solve D with segment tree.

your query function works with O(26*log(26)*log(N)*Q) it is about 10^9 operations

You merge 2 sets in each node of your tree. It's something like O(q * n * log(N) * log(N))

dont use unordered map without custom hash or else you'll get tle.

Also, lot of problems are tagged something but there is no reference solution (in tutorial) which uses that tag. So, the tag is more or less useless.

its becuas dp is double important :)))

fixed & thx

Your rating won't change unless you submit any solutions.

ceil returns a double/float. you can make that:

int ceil2(int a, int b){
    return a / b + !!(a % b);
}

https://en.wikipedia.org/wiki/IEEE_754

U can use a well known trick to divide n by k rounding up. This value is a (n + k — 1) / k.

This is why unordered_map may exceed the time limit in some cases.

Few useful functions of TreeSet that resembles binary search: tree.floor(int), tree.ceil(int), tree.higher(int), tree.lower(int)

My solution with IT: https://codeforces.me/contest/1234/submission/61659385 350ms

My solution with IT: https://codeforces.me/contest/1234/submission/61659385 350ms

Your solution works O(26*log(26)*log(N)*Q) it is about 10^9 operations

your solution working in O(q*logn*n*log27), may be that so, because each of node you merged 2 two set. sorry my bad english

He very humorous =))

dependent on data type, umap is faster than map when you can sort large arr

Yeah when there is alot of collisons then unordered_map isn't O(1). However ordered_map is always o(1)

First you can start with the idea that there is only a single path that can allow flow of water from (1,1) to (2,n+1),because if you receive water from one side of pipe, another end can only face a fixed direction if you are getting water from fixed and unique direction (except for the case when you have tilted pipe and receive water from left side to that pipe, so it could continue flow in up or down direction ). So simulate that flow till either an out of bounds occur, otherwise if you reach (2,n+1) before, your answer is "YES", otherwise "NO". Hope it helps