Bitmask DP doing some magic

Правка en2, от micklepru, 2015-07-25 01:11:12

Hello, Codeforces!

I ask you to help me understanding how does the algorithm work.

Here is the problem. Basically we have two integers N and M, we need to find how many numbers obtained by rearranging digits of N are divisible by M.

So naive solution is just go through all permutations of digits of N and check all of them. That gives us N! operations, which is unacceptably slow. Intuitively, we must go through all permutations anyway. And I am very confused about how do we not.

We use bitmask DP: dp[i][j] — amount of numbers, obtained by permutation some digits of N(specified by bitmask i), and remainder after diving it by M is j. For example N=12345, dp[01101][2] — amount of numbers, constructed with 2,3,5 (such as 352), and after division by M remainder is 2 (M=50, 352%50==2). For each bitmask i we iterate(variable k) through all unused yet digits((i & 1<<k)==0) and add it to the end of every permutation(i | 1<<k). Old permutations gave remainder j, now it will be (j*10+N[k])%M, where N[k] is k-th digit of N. So we increase dp[i | 1<<k][(j*10+N[k])%M] by dp[i][j].

Here is the code to make it easier: 12205724 Note: timesRepeat is used to eliminate duplicate permutations regarding repeated digits in N. (i||N[k]) is to get rid of leading zeros

Using DP, we descend from N! to (2^N)*N, which gives us accepted.

Consider the solution: seems like we iterate through every permutation. For example, bitmask 0110 can be obtained as 0100 adding third digit to the end or 0010 adding second digit to the end. I cannot understand how are we doing less work, but logically we go over every permutation.

Maybe it is stupid question, but I am really confused. Please help me to sort it out and tell when do we use such method. Can we use it to any kind of permutations?

Теги dp, bitmask, permutations, brute force

История

 
 
 
 
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  Rev. Язык Кто Когда Δ Комментарий
ru2 Русский micklepru 2015-07-25 15:07:53 117
en3 Английский micklepru 2015-07-25 15:05:11 100 Tiny change: '#### UPD:\n_Thank you' -
en2 Английский micklepru 2015-07-25 01:11:12 11 Tiny change: 'nused yet numbers(`(i & 1<' -> 'nused yet digits(`(i & 1<'
ru1 Русский micklepru 2015-07-25 00:47:57 2150 Первая редакция перевода на Русский
en1 Английский micklepru 2015-07-25 00:28:07 1940 Initial revision (published)