А что есть v в задаче Д?

D, ответ поделить на количество вхождений каждой цифры — здесь должно быть "на факториалы количества", правильно?

Можно в разбор добавить так же ссылки на авторские исходники? Так будет проще понять сложные моменты. К примеру, Е без исходника как-то очень трудно воспринимается.

dp[i or (1 shl (j-1)),(x*10+a[j]) mod m] := dp[i or (1 shl (j-1)),(x*10+a[j]) mod m] + dp[i,x];

И все сразу стало понятно...

When will the editorials be published ?

Не понял решения D,E. Объясните, плз.

А в задаче С на FPC достаточно было склеить строку-ответ и вывести ее в самом конце)

Возник вопрос по задаче D. Есть две почти одинаковые отправки: 5991586 и 5991163. Отличие одно: в первой — массив w[2^18][100], во второй — w[100][2^18] (соответственно меняется подсчет). В первом случае решение проходит за 500 с небольшим мс, во втором — лимит по времени не на самом тяжелом тесте. Почему такая большая разница (как минимум, в 8 раз)?

I know the problem D should be dynamic programming, but I still don't know how to solve this problem

I know the problem D should be dynamic programming, but I still don't know how to solve this problem

I think that you should use

Unable to parse markup [type=CF_TEX]

I think that you should use

Unable to parse markup [type=CF_TEX]

It would be good if authors solution is in C/C++ or java.

В "B" создал строку из 1-иц и 0-ов, где нолики — незанято. Тогда ответ составил количество успешных совпадений при жадном поиске c регулярным выражением (regexp) по всей строке (global) — /00|0/g и /0/g. Perl код 5990466.

For problem D can you tell me what 'mask of reshuffle' means? sorry i don't get the solution clearly

Sorry, but I couldn't understand the editorial for Problem D. Others are also very short. Can you elaborate a little? Is there any similar problem at UVA or TopCoder or any other online judge? If you know please give the numbers if possible,

In the end we must answer to divide by the factorial number of occurrences of each digit.

What does this even mean ? The english is not correct and I don't understand why we must divide de result ?

For problem D,

I have converted the dp equation to c++

i goes from 0 to 2^l-1
	j goes from 0 to l-1
		x goes from 0 to m-1
			dp [i | (1 << (j-1))] [(x*10+a[j]) mod m] += dp[i][x];

Is my understanding true that dp[i][x] represent the answer for the number whose elements are selected by the i bit mask with divisor as x. In other words it is the answer to the number given by the selected digits and when the divisor is x.

If someone can explain the dp equation in simple terms, that would be a great help. I can only understand that the number with one digit less is somehow related to the number with one digit more.

It's ironic how some problem setters are so smart at coming up with problems, but so lousy at explaining solutions.

UPD: Sorry, my bad, I didn't read problem statement correctly.

I don't know if anyone will pay attention to my comment now, but I was doing Problem C while practicing and found that the alternative solution is not being considered by the judge.

In case 5, ie. 3 4, my code outputs 1101101 but the judge declares it's wrong, citing the Jury's solution as 1010101 and message as wrong answer. The number of ones should be 4, but found 5; The number of zeros should be 3, but found 2;

My submission is here: 11114404

After re-reading the question several times, I'm quite sure my answer satisfies all constraints mentioned in problem. 1101101 doesn't have more than 2 adj. 1 and no more than 1 adj. 0.

I request the author, EG0R to kindly look into this problem and resolve this issue(and in case it's my own mistake, point it out.). :)

Ladies and gentlemen, is this the highest dimensional dp ever written ? http://codeforces.com/contest/401/submission/22059230

For problem D , my submission get time limit on test 35 and I can't find how to optimize my solution.

This is my submission 36532811 . Any help please !

I think the implementation for C is quite cumbersome.
Let $$$n_0$$$: the number of 0, $$$n_1$$$ the number of 1.
$$$n_1$$$ must be in constraint $$$n_0-1 \leq n_1 \leq 2n_0+2$$$, otherwise there is no solution.
I will solve for range: $$$n_0 \leq n_1 \leq 2n_0$$$, (you should see my code to understand correctly)
Let $$$a$$$ is the number of patterns '110', $$$b$$$ is the number of patterns '10', so:

We are solving in range $$$n_0 \leq n_1 \leq 2n_0$$$, therefore $$$a, b \leq 0$$$. With $$$a, b$$$ after solving two equations. We construct a string answer following the greedy strategy. Up to now, the string answer always start with '1' and end with '0'.
I add some padding characters to get answer.
if $$$n_1 = n_0 - 1$$$, add '0' to start of above string.
if $$$n_1 > 2n_0$$$, add $$$(2n_0 - n_1)$$$ '1' to the end of string.
My solution

c ..i think there would be a better soln than this..31ms i got!!!

void solve() { ll n,m; cin>>n>>m; if(m>2*n+2 || n>m+1) cout<<"-1"<<endl; else if(n==m) { string s=""; for(ll i=1;i<=m;i++) s+="10"; cout<<s<<endl; return; } else if(n==m+1) { string s="0"; for(ll i=1;i<=m;i++) s+="10"; cout<<s<<endl; return;

}
else if(n+1==m)
{
    string s="";
           for(ll i=1;i<=n;i++)
                      s+="10";
    s+="1";
                  cout<<s<<endl; return;
}
else
{
    string s="";
    ll used=min3((m-n),n,m);
    ll rem_n=n-used,rem_m=m-(2*(used));
           n=rem_n; m=rem_m;

     if(n==m+1)
      {
          s="0";
          for(ll i=1;i<=m;i++)
              s+="10";
          for(ll i=1;i<=used;i++)
            s+="110";
          cout<<s<<endl; return;
      }
      else if(m==n+1)
      {
          for(ll i=1;i<=used;i++)
                s+="110";
          s+="1";
                 for(ll i=1;i<=n;i++)
                     s+="10";
          cout<<s<<endl; return;
         // m=0; n=0;
      }
      else if(m==n)
      {
          for(ll i=1;i<=used;i++)
                s+="110";

               for(ll i=1;i<=m;i++)
                   s+="10";
           cout<<s<<endl; return;
      }

    for(ll i=1;i<=used;i++)
                       s+="110";
     //cout<<s<<endl; return;

    while(m--)
        s+="1";

    cout<<s<<endl;
}

}

can anyone explain how to solve the D problem if I am using normal masking it is giving me tle because 2^18 for mask and 100 for the reminder and for each case we will check at all digit that is 18 so I think there will better solution or how should I optimize this.

" then we must start form one and we derive the ones and zeros in one, but in the end must be one. And if we in the end and we have ones, then we try to stick to one unit so that we have. For example, we have 10101 and two ones, after we have 110101 and one ones, and then we have 1101101."

Crap explanation. Better learn some english first before writing editorials.