Submission #1027614

#	Time	Username	Problem	Language	Result	Execution time	Memory
1027614		caterpillow	"The Lyuboyn" code (IZhO19_lyuboyn)	C++17	3 / 100	182 ms	34404 KiB

lyuboyn

#include <bits/stdc++.h>

using namespace std;

using db = long double;
using ll = long long;
using pl = pair<ll, ll>;
using pi = pair<int, int>;
#define vt vector
#define f first
#define s second
#define pb push_back
#define all(x) x.begin(), x.end() 
#define size(x) ((int) (x).size())
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define F0R(i, b) FOR (i, 0, b)
#define endl '\n'
const ll INF = 1e18;
const int inf = 1e9;

template<typename Tuple, size_t... Is>
void print_tuple(const Tuple& t, index_sequence<Is...>) {
    ((cerr << (Is == 0 ? "{" : ", ") << get<Is>(t)), ...);
    cerr << "}";
}

template<typename... Args>
ostream& operator<<(ostream& os, const tuple<Args...>& t) {
    print_tuple(t, index_sequence_for<Args...>{});
    return os;
}

ostream& operator<<(ostream& os, string& s) {
    for (char c : s) os << c; 
    return os;
}

template<template<typename> class Container, typename T>
ostream& operator<<(ostream& os, Container<T> o) {
    os << "{"; 
    int g = size(o); 
    for (auto i : o) os << i << ((--g) == 0 ? "" : ", "); 
    os << "}";
    return os;
}

template<typename T, typename V>
ostream& operator<<(ostream& os, const pair<T, V> p) {
    os << "{" << p.f << ", " << p.s << "}";
    return os;
}

template <typename T, typename... V>
void printer(const char *names, T &&head, V &&...tail) {
    int i = 0;
    while (names[i] != '\0' && names[i] != ',') i++;
    constexpr bool is_str = is_same_v<decay_t<T>, const char*>;
    if constexpr (is_str) cerr << head; // ignore directly passed strings
    else cerr.write(names, i) << " = " << head; 
    if constexpr (sizeof...(tail)) cerr << (is_str ? "" : ","), printer(names + i + 1, tail...);
    else cerr << endl;
}

#ifdef LOCAL
#define dbg(...) std::cerr << __LINE__ << ": ", printer(#__VA_ARGS__, __VA_ARGS__)
#else
#define dbg(x...)
#define cerr if (0) std::cerr
#endif

/*

each step is xoring prev number with a new mask of popcount k
s is irrelevant

if k is even, then the parity will never change, and hence its impossible
this is a necessary condition, and since we do info and not math we just assume its also necessary 

*/

int n, k, t, s;
vt<bool> seen;
vt<int> transitions;

void prnt(int x) {
    ROF (i, 0, n) {
        cout << (1 & (x >> i));
    }
    cout << endl;
}

bool dfs(int cur, int d = 0) {
    if (d == (1 << n) - 1) {
        if (__builtin_popcount(cur) == k) {
            prnt(cur);
            return true;
        } else return false;
    }
    seen[cur] = true;
    for (int t : transitions) {
        if (!seen[cur ^ t]) {
            if (dfs(cur ^ t, d + 1)) {
                prnt(cur ^ s);
                return true;
            }
        }
    }
    seen[cur] = false;
    return false;
}

main() {
    cin.tie(0)->sync_with_stdio(0);
    
    cin >> n >> k >> t;
    if (k % 2 == 0) {
        cout << "-1\n";
        return 0;
    }
    string st; cin >> st;
    F0R (i, size(st)) {
        s <<= 1;
        s += (st[i] == '1');
    }
    cout << (1 << n) << endl;
    F0R (i, 1 << n) {
        if (__builtin_popcount(i) == k) {
            transitions.pb(i);
        }
    }
    seen.resize(1 << n);
    assert(dfs(0));
}

Compilation message (stderr)

lyuboyn.cpp:113:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
  113 | main() {
      | ^~~~

• Subtask 1: 0 / 0
• Subtask 2: 0 / 5
• Subtask 3: 3 / 3
• Subtask 4: 0 / 11
• Subtask 5: 0 / 15
• Subtask 6: 0 / 18
• Subtask 7: 0 / 31
• Subtask 8: 0 / 17