Submission #414658

#	Time	Username	Problem	Language	Result	Execution time	Memory
414658		JerryLiu06	Snake Escaping (JOI18_snake_escaping)	C++17	100 / 100	1132 ms	43312 KiB

snake_escaping

#include <bits/stdc++.h>
 
using namespace std;
 
using ll = long long;
using db = long double;
using str = string;
 
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
 
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
 
#define mp make_pair
#define f first
#define s second

#define sz(x) (int)(x).size()
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define sor(x) sort(all(x))
#define ft front()
#define bk back()
#define pb push_back
#define pf push_front
 
#define lb lower_bound
#define ub upper_bound
 
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define R0F(i, a) ROF(i, 0, a)
#define EACH(a, x) for (auto& a : x)

const int MOD = 1e9 + 7;
const int MX = 20;
const ll INF = 1e18;

int L, Q; string S; int A, B, C; // A = '0', B = '1', C = '?' 

int cntA, cntB, cntC;

int cnt[(1 << MX)], sup[(1 << MX)], sub[(1 << MX)]; // sup - SUM_{mask \in i}, sub - SUM_{i \in mask}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);

    cin >> L >> Q >> S; F0R(i, (1 << L)) sup[i] = sub[i] = S[i] - '0', cnt[i] = __builtin_popcount(i);

    F0R(i, L) F0R(j, (1 << L)) {
        if (!(j & (1 << i))) sup[j] += sup[j ^ (1 << i)];
        if ((j & (1 << i)))  sub[j] += sub[j ^ (1 << i)];
    }
    F0R(i, Q) {
        string cur; cin >> cur; A = B = C = cntA = cntB = cntC = 0; ll ans = 0;

        F0R(j, L) {
            if (cur[j] == '0') A ^= (1 << (L - j - 1)), cntA++;
            if (cur[j] == '1') B ^= (1 << (L - j - 1)), cntB++;
            if (cur[j] == '?') C ^= (1 << (L - j - 1)), cntC++;
        }
        if (cntA <= L / 3) {
            for (int X = A; X > 0; X = (X - 1) & A) {
                ans += (1 - 2 * (cnt[X] & 1)) * sup[B | X];
            }
            ans += sup[B]; cout << ans << "\n";
        }
        else if (cntB <= L / 3) {
            for (int X = B; X > 0; X = (X - 1) & B) {
                ans += (1 - 2 * (cnt[X] & 1)) * sub[C | X];
            }
            ans += sub[C]; cout << abs(ans) << "\n";
        }
        else { for (int X = C; X > 0; X = (X - 1) & C) ans += S[B | X] - '0'; ans += S[B] - '0'; cout << ans << "\n"; }
    }
}

• Subtask 1: 5 / 5
• Subtask 2: 7 / 7
• Subtask 3: 10 / 10
• Subtask 4: 53 / 53
• Subtask 5: 25 / 25