Hope, history will not repeat because in this contest, there are 22 problem setters and testers.

educate yourself if you want high rating!

No you're not, i also can't open.

I think after this round everything will be ok.

maybe it's just to prevent delay submission during the contest, which participant complaint for each contest lul

Some reason is problem with codeforces api.

Can it be because the API is not working which in turn is to prevent us from seeing questions done by testers.

I'm sorry dude, but it is how cf works in community...
You get tones of downvotes if your rating's low...
You can't hold a contest because your rating's low...
There might be some issues if you hold a contest and your rating's low...
I don't have an idea how to improve my rating after years being on this website, Nor to be better in community. Hope someone help me in some new ways... . And something else, please think about it before starting downvote with millions of account(fake or real :()

It was because you solution was not fast enough to pass the remaining system tests ! during contest only pretests are tested that's why it shows pretest passed ! hope you got it

During the contest, your code is tested on a subset of the final tests, called PRETESTS. After you have passed the pretests, you can lock your solution and then view other participants' codes and try to hack them. Others who have locked their solutions can view your code irrespective of whether or not you have locked your code. Once the coding phase is over, your solutions will be tested on the entire set of tests. This phase is called system testing.

It is the system testing, in the round, the judge runs only about 15 cases on your code, if your code passed you got PREtests passed. And after the contest, they run around another 100 test on your code and give you your final verdicts. You can also learn about hacking a solution.

Why does people downVote your comments without any reason (maybe even without reading your comment) ?

Can you share your solution to E

E is pretty easy. It's a standard fenwick tree problem. I'm ashamed I couldn't solve C though T_T

Disqualified. However, it's better to write such messages in "ask a question" during a contest.

jealous !

1 1 1 1 1

lel. you mean sort(b+1, b+1+k)? or is it not Office Keys problem?

I failed bin search because of l=1=)))

3
10 8 8

Answer is "NO".

there were many!! someone in my room has passesd pretests with this soluion http://codeforces.me/contest/831/submission/28522407 !!

2 1 2 or 2 2 1

Eagerly waiting for the replies:) Can someone explain there idea please

D is very simple question of basic DP. sort all men and keys on the basis of their positions and then u have dp state as dp[men][key] = min(max(time taken to that men after picking that key , dp[men+1][key+1]) ,dp[men][key+1]). and after all recursive calls ...your final answer is dp[0][0].

I solved D using binary search + Maximum matching

For problem D:

I guess that the chosen key should be a consecutive n keys segment.. I don't know whether it will pass, but it seems pretty simple..

I solved D as follows: It's optimal to use a set of keys that are neighbours. Sort the keys, sort the people. Let the rightmost person go get the rightmost key, etc. Try every key as rightmost key.

Lets say your path starts at 0. Create array for its values (adding a_i). Now make some path value to b_0.

You know b[0] must be after some judge i, so check for every position i, if this could work in linear time

let us get the prefix sum for all A's, now sort them. then get the value of all B's and sort them. now let us say the max value in B is x, now for x to be at position i, the initial value will be x — pref_sum[i]. then using this initial value generate the the array and check whether all numbers come in b.

Yeah. I got TLE on pretest 12 :(

You can preprocess it?

There's a k^2 log k solution though

Pretests are so weak that my solution without sorting passed. And even wasn't hacked :( RIP rating

For Div 1A / Div 2D Not only pretests, I saw a wrong solution passing the system test.

    ll res=INT_MAX;
    sort(a,a+n);
    sort(b,b+k);
    for(int i=0;i<k;i++)
    {
        ll ans=0;
        for(int j=0;j<n;j++)
        {
            ll tmp = abs(a[j]-b[i+j]) + abs(b[i+j]-p);
            ans=max(ans,tmp);
        }
        res=min(res,ans);
    }

It can be seen that for array b index will reach to n+k whereas b can have max index possible till k.

Link to Code

I used bs + max matching