IDoNotKnowAnything's blog

Need help in CS Academy Question : "And Closure"
By IDoNotKnowAnything, history, 8 years ago, In English

Can someone help me in solving this question here.I had tried to solve it for hours but could not solve it.Later I watched the editorial but I am unable to get it.I understood the basic idea of role of bit being set 0 .But Still I am unable how this fact is incorporated in the solution . Can someone please help me with the problem ???? Thanks in advance and have a good day :)

  • Vote: I like it
  • +4
  • Vote: I do not like it

»
8 years ago, # |
← Rev. 5   Vote: I like it +9 Vote: I do not like it

I loved this problem. It was beautiful. I understood it in the following fashion:

  1. It tells how closely we can make x. In beginning for all x, dp[x] = 1111...11 Then we set dp[x] = x, for . If you do not understand what I am doing, continue reading the example mentioned below.

Eg: Consider number 12 i.e (1100)2. Let at some stage dp[(1100)2] = 1101, then it means we are able to turn off the 3rd bit but 4th bit can't be set to 0 right now.
Read the following very carefully.
dp[(1100)2] = AND(xi), where xi is in array and contains 12 i.e xi&12 = (1100)2.
If we perform AND operation on all such , and if still dp[12]! = 12, means 12 can't be formed.

  1. If ith bit is set in x, then lets turn it off and the new number be .
    Then dp[y] &= dp[x], What does this do?
    Clearly x contains y, and dp[x] tells us what bits we can turn off to reach as close to x. It's interesting to note that all those bits can also be turned off while trying to form y. So we perform dp[y] &=dp[x], and try to transfer all those 0 bits from dp[x] to dp[y]. Since x contains y, doing this operation will only help dp[y] to have 0s in appropriate bit positions and reach closer to y.
    If you will look carefully, dp[x] is a number obtained by AND operation on all such numbers that contain x, and are present in array. Since x contains y, and we perform the AND operation, all those numbers are added to the AND collection of dp[y].

Now, just do the above DP operation, iterating from 106 to 1. At end, ans is number of x such that dp[x] = x Code

→ Reply
  • »
    »
    8 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    Got it .Thanks for reply .So basically we are trying to supply all possible zeroes that we have in a number x to all numbers y that can be formed from it through & operations .Thanks :)

    → Reply
»
8 years ago, # |
  Vote: I like it +7 Vote: I do not like it

I have another solution which is a bit different from the editorial's solution.

First using Sum over Subsets DP (Supersets in this case) , for each find the number of elements which are supersets of , lets call it , then let , on this do an inverse sum over subsets dp to find number of subsets with exactly equal to . This not only gives the values possible possible but also their count.

Lastly this might overflow so use 2 modulos to hash it.

Code.

Also the last Div1 D is a bit similar (just change base 2 to base 10).

→ Reply
  • »
    »
    8 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Thanks for the alternative approach but first I have to read SOS DP(I do not know that too :(

    → Reply
  • »
    »
    7 years ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    Your solution as well as the article you provided are very useful. I am the author of this problem and my initial solution was with a variation of FFT(Link). Thank you for sharing this approach with us. I did not know about SOS dp and now everything seems clearer. Your solution is also very ingenious.

    → Reply