제출 #1224694

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1224694		LIA	Unscrambling a Messy Bug (IOI16_messy)	C++17	0 / 100	0 ms	328 KiB

messy

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef tuple <ll,ll,ll> plll;
typedef vector <plll> vplll;
typedef pair <ll,ll> pll;
typedef vector <ll> vll;
typedef vector <pll> vpll;
typedef vector <vector <pll>> vvpll;
typedef vector <vector <ll>> vvll;
typedef vector <bool> vb;
typedef vector <vector <bool>> vvb;
#define loop(i, s, e) for (ll i = (s); i < (e); ++i)
#define loopr(i, e, s) for (ll i = (e)-1; i >= (s); --i)
#define all(a) a.begin(), a.end()
const ll inf = 1e9 + 7;
 
#include "messy.h"

vector<int> restore_permutation(int n, int w, int r) {
    int b = 0;
    while ((1<<b) < n) ++b;
    vector<string> S(b), T;
    for (int k = 0; k < b; ++k) {
        S[k].resize(n);
        for (int i = 0; i < n; ++i)
            S[k][i] = ((i>>k)&1) ? '1' : '0';
        add_element(S[k]);
    }
    compile_set();
    for (int mask = 0; mask < (1<<n); ++mask) {
        string t(n,'0');
        for (int i = 0; i < n; ++i)
            if (mask & (1<<(n-1-i)))
                t[i] = '1';
        if (check_element(t))
            T.push_back(t);
    }
    vector<int> p(n);
    vector<int> idx(b);
    iota(idx.begin(), idx.end(), 0);
    do {
        vector<int> q(n,0), cnt(n,0);
        for (int j = 0; j < n; ++j) {
            int v = 0;
            for (int k = 0; k < b; ++k)
                if (T[idx[k]][j]=='1')
                    v |= 1<<k;
            if (v >= n) { v = -1; break; }
            q[j]=v; ++cnt[v];
        }
        bool ok = true;
        for (int i = 0; i < n; ++i)
            if (cnt[i]!=1) { ok = false; break; }
        if (ok) { p = q; break; }
    } while (next_permutation(idx.begin(), idx.end()));
    return p;
}

컴파일 시 표준 에러 (stderr) 메시지

messy.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~
messy_c.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~

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