Submission #597621

#TimeUsernameProblemLanguageResultExecution timeMemory
597621jophyyjhMartian DNA (IOI16_dna)C++14
100 / 100
12 ms384 KiB
/** * OK, so we have a binary str of len n. Each time we're allowed to give a str and * whether the str is a substr of the original binary str is returned. The task is * to determine the entire str under a certain num of interactions. * * Well it looks to me that the num of steps shall be O(n). Well, i think i've got * a solution by extending the current str on either side. 2n steps is now the * maximum. Using randomized algo, we can prove that the expected num of * interactions <= 1.5n, which is still too much, and whether the interactor is * adaptive remains unknown. * * We want to further lower the num of interactions. Can we directly determine the * first char? Well i guess the goal is to find the suffix (or equivalently the * prefix). In the algo above, we extend our str on the right side until it can * no longer be extended; after this, we can just extend it on the left side, * without fearing that a "true" response corresponds to a substr which is not a * suffix (shifted pos). In other words, we wish to find a substr such that: * make_test((substr)0), make_test((substr)1) are all false. * Hmm, this doesn't seem to work. My second solution is quite interesting. We * begin by testing 0, if 0 isn't there we add 1 to our current str. Each time, we * test a char and add it if it's a true, otherwise we add the other char. So, we * know that now only a prefix of our str is actually in the original str, so * there's where we can use binary search. * ------------------------------ After Some Hints ------------------------------ * We try to improve the process of finding a suffix. The "search to the RHS + * binary search" method works, but we need to be careful about WHEN to start our * binary search. We don't have to ask 0/1 randomly, but can instead default to * asking 1. Hmm, when we've exceeded the RHS, we naturally get a series of * "false", meaning that we would think that a series of 0000...0 is added. Now * comes the ingenious part. We first use binary search to find the longest * contigous chain with all 0s (suppose the len is k). Therefore, we use at most * k+1 more steps ((k+1) consecutive "false"s), but then we determined the k 0s in * log(n) steps. So we have: (n-k)+log(n)+(k+1)+log(n) ~ n + 2log(n). * * Number of steps: n + 2log(n) * Implementation 2 */ #include <bits/stdc++.h> #include "dna.h" std::string analyse(int n, int T) { int longest_0 = 0; for (int step = n / 2 + 1; step >= 1; step /= 2) { while (longest_0 + step <= n && make_test(std::string(longest_0 + step, '0'))) longest_0 += step; } std::string current(longest_0, '0'); for (int consec_zero = 0; consec_zero <= longest_0; ) { if (make_test(current + '1')) current += '1', consec_zero = 0; else current += '0', consec_zero++; } int len = 0; for (int step = n / 2 + 1; step >= 1; step /= 2) { while (len + step <= std::min(int(current.size()), n) && make_test(current.substr(0, len + step))) { len += step; } } current = current.substr(0, len); while (int(current.length()) < n) { if (make_test('1' + current)) current.insert(0, 1, '1'); else current.insert(0, 1, '0'); } return current; }

Compilation message (stderr)

grader.cpp: In function 'bool make_test(std::string)':
grader.cpp:14:20: warning: comparison of integer expressions of different signedness: 'int' and 'std::__cxx11::basic_string<char>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   14 |  for (int i = 0; i < p.size(); i++) {
      |                  ~~^~~~~~~~~~
grader.cpp:23:20: warning: comparison of integer expressions of different signedness: 'int' and 'std::__cxx11::basic_string<char>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   23 |  for (int i = 1; i <= ss.size(); i++) {
      |                  ~~^~~~~~~~~~~~
grader.cpp:28:13: warning: comparison of integer expressions of different signedness: '__gnu_cxx::__alloc_traits<std::allocator<int>, int>::value_type' {aka 'int'} and 'std::__cxx11::basic_string<char>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   28 |   if (pr[i] == p.size()) {
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