Блог пользователя Shafaet

HackerRank HourRank 19
Автор Shafaet, история, 7 лет назад, По-английски

I invite you to take part in HackerRank HourRank 19 starting on April 2nd 2017. The contest duration is 1 hours. There will be three algorithmic problems of various difficulty.

The chief author of this round is torquecode. The chief tester is pkacprzak. Thanks to Piotr Gajowiak and Wild_Hamster for pre-solving the problems and giving valuable opinions.

The problems will have subtasks to make them interesting for everyone. I strongly recommend to read all the problems.

The contestants are ranked by score. If two contestants get the same score, the person who reached the score first is ranked higher.

Happy Coding!

  • Проголосовать: нравится
  • +16
  • Проголосовать: не нравится

»
7 лет назад, # |
  Проголосовать: нравится +3 Проголосовать: не нравится

Wow a 1 hour event again.......Thanks for the info :0 Hoping for better rating this time .

→ Ответить
»
7 лет назад, # |
  Проголосовать: нравится +5 Проголосовать: не нравится

For the last question,instead of removing each edge and calculating the diameters of the 2 trees obtained..it would be sufficient to just remove edges incident on the center of the tree and which are part of diameter(ie only 2 edges). Although I'm not aware of the proof,I'm quite sure that this should also be correct.

→ Ответить
»
7 лет назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

Third problem is similar to a COCI problem which I think is the reason why people solved it so quickly.

→ Ответить
»
7 лет назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

Can someone explain how to solve the second problem.

→ Ответить
  • »
    »
    7 лет назад, # ^ |
      Проголосовать: нравится 0 Проголосовать: не нравится

    Read Nim

    → Ответить
  • »
    »
    7 лет назад, # ^ |
      Проголосовать: нравится +5 Проголосовать: не нравится 
    let indexes are 0..n-1.

1.   s[b] XOR s[b+1] XOR ... XOR s[e] == s[0] XOR s[1] XOR ... XOR s[n-1] = stotalxor = const.
2.   s[b] XOR s[b+1] XOR ... XOR s[e] == ( s[0] XOR s[2] .. XOR s[b-1])  XOR ( s[0] ... s[e] ) =
     =  v[b] XOR v[e+1]
3.  for each 0 <= e < n   need find how many b <= e, that v[b] XOR v[e+1] == stotalxor.
    or  v[b] = stotalxor XOR v[e+1].

where v[i] = s[0] xor s[2] ... s[i-1] .
v[i] - easyliy computeted, because  v[0] = 0,  v[i] = v[i-1] xor s[i-1], for i > 0.

  and v[i] <= 2^17.

Let  xorcnt[ i ] - number of that j,   v[j] = i.

xorcnt[i] - also easyly computed.

xorcnt[v[0]] = 1;
ans = 0;
for(int i= 1; i <= n; ++i){  ans += xorcnt[ v[ i ] ^ sxortotal ];  xorcnt[ v[i] ] ++; }
cout << ans;
    → Ответить
    • »
      »
      »
      7 лет назад, # ^ |
        Проголосовать: нравится 0 Проголосовать: не нравится

      well if i got you right finding all combinations of v[b] ^ v[e] == stotalxor is the main idea of solution?

      → Ответить
      • »
        »
        »
        »
        7 лет назад, # ^ |
        ← Rev. 3   Проголосовать: нравится 0 Проголосовать: не нравится

        Nim game: first player can win if only if xor sum s[] elements is not zero.

        But, our case need count winning situations of second player, so need achieve xor sum of s[] elements to zero, after remove s[b]...s[e] elements.

        i.e.

        s[0] xor s[1] xor ... xor s[b-1] xor s[e+1] xor ... s[n-1] == 0.

        add xor to both sides to (s[b] xor s[b+1] xor .. s[e]) — gives

        stotalxor = xor_sum(s[i], i= 0..n-1) = 0 xor xor_sum(s[j], j = b..e) = xor_sum(s[j], b..e).

        xor_sum(s[j], j = b..e) = xor_sum(s[j], j = 0..b-1) xor xor_sum(s[j], j = 0..e)

        Let v[k] = xor_sum(s[j], j =0..k-1);

        so need find that (b,e) pairs (0 <= b <= e < n ), that v[b] ^ v[e+1] == stotalxor.

        if we know e, so can calculate number of b, that 0 <= b <= e, and v[b] ^ v[e+1] == stotalxor. because,

        v[b] = stotalxor ^ v[e+1].  if we store number of v[i] ,  0<=i<=e  in xocrcnt[] array, number

        of b is xorcnt[ stotalxor ^ v[e+1] ].

        → Ответить