제출 #402112

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
402112		blue	Unscrambling a Messy Bug (IOI16_messy)	C++17	20 / 100	3 ms	460 KiB

messy

#include "messy.h"
#include <string>
#include <vector>
using namespace std;

/*
Let n = 8
First, we can insert

10000000                all i such that p[i] <= n/2
01000000
00100000
00010000

Now, we can query
10000000
...
00000001
where 1 is only at position i
The positions i where the query is true, satisfy the following:
    p_inv[i] <= n/2

10001111
01001111
11111000
11110100

10111111
11101111
11111011
11111110



From these, we can identify all i such that (p[i] <= n/2).


Let later sequence = b and input sequence = a
a[i] == 1
b[j] == 1
j = p[i]

We identify p[i]
X[i] <= p[i] <= Y[i]

for X = n, n/2 ... 2
We partition the set {i} into two groups, those with p_inv[i] % n <= n/2,
and those with p_inv[i] % n > n/2

*/


int N;
vector<int> p_inv(128, 0);

void find_p_inv(int a, int b, vector<int> goodcells)
//goodcells contains all the elements with p_inv in the range a..b
{
    if(a == b)
    {
        p_inv[goodcells[0]] = a;
        return;
    }

    string base_query(N, '1');
    for(int g: goodcells)
        base_query[g] = '0';


    vector<int> goodcells_1, goodcells_2;

    for(int g: goodcells)
    {
        string query = base_query;
        query[g] = '1';

        if(check_element(query))
            goodcells_1.push_back(g);
        else
            goodcells_2.push_back(g);
    }

    find_p_inv(a, (a+b)/2, goodcells_1);
    find_p_inv((a+b)/2+1, b, goodcells_2);
}

vector<int> restore_permutation(int n, int w, int r)
{
    N = n;
    for(int g = n; g >= 2; g /= 2) //size of gap
    {
        for(int i = 0; i < n; i += g) //starting point of gap
        {
            for(int o = 0; o < g/2; o++)
            {
                string s(n, '1');
                for(int j = i; j < i+g; j++)
                    if(j-i != o)
                        s[j] = '0';
                add_element(s);
            }
        }
    }

    compile_set();

    // cerr << n << '\n';
    vector<int> goodcells;
    for(int i = 0; i < n; i++) goodcells.push_back(i);
    find_p_inv(0, n-1, goodcells);

    vector<int> p(n);
    for(int i = 0; i < n; i++)
        p[p_inv[i]] = i;

    return p;
}

• Subtask 1: 20 / 20
• Subtask 2: 0 / 18
• Subtask 3: 0 / 11
• Subtask 4: 0 / 21
• Subtask 5: 0 / 30