Submission #597617

#	Time	Username	Problem	Language	Result	Execution time	Memory
597617		jophyyjh	Martian DNA (IOI16_dna)	C++14	0 / 100	15 ms	384 KiB

dna

/**
 * 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 = -1;
    for (int step = n / 2 + 1; step >= 1; step /= 2) {
        while (len + step < 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()) {

• Subtask 1: 0 / 11
• Subtask 2: 0 / 25
• Subtask 3: 0 / 64