Submission #396894

#	Time	Username	Problem	Language	Result	Execution time	Memory
396894		Benq	Snake Escaping (JOI18_snake_escaping)	C++17	100 / 100	1130 ms	21424 KiB

snake_escaping

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
 
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
 
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
 
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 1<<20;
 
int L,Q, dp[2][MX];
short pc[MX], val[MX];
 
void init() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> L >> Q;
    FOR(i,1,MX) pc[i] = pc[i-(i&-i)]+1;
    string S; cin >> S;
    F0R(i,1<<L) dp[1][i] = dp[0][(1<<L)-1-i] = val[i] = S[i]-'0';
    
    F0R(i,L) F0R(j,1<<L) if (j&(1<<i)) 
        F0R(k,2) dp[k][j] += dp[k][j^(1<<i)];
}
 
int solve(const string& T) {
    int a = 0, b = 0, c = 0;
    F0R(i,sz(T)) {
        switch(T[sz(T)-1-i]) {
            case '0':
                a ^= 1<<i;
                break;
            case '1':
                b ^= 1<<i;
                break;
            default:
                c ^= 1<<i;
                break;
        }
    }
    
    int ans = 0;
    int mn = min(min(pc[a],pc[b]),pc[c]);
    if (pc[a] == mn) {
        for (int m = a; ; m = (m-1)&a) {
            if ((pc[a]-pc[m])&1) ans -= dp[0][m|c];
            else ans += dp[0][m|c];
            if (m == 0) break;
        }
    } else if (pc[b] == mn) {
        for (int m = b; ; m = (m-1)&b) {
            if ((pc[b]-pc[m])&1) ans -= dp[1][m|c];
            else ans += dp[1][m|c];
            if (m == 0) break;
        }
    } else {
        for (int m = c; ; m = (m-1)&c) {
            ans += val[m|b];
            if (m == 0) break;
        }
    }
    
    return ans;
}
 
int main() {
    init();
    F0R(i,Q) {
        string T; cin >> T;
        cout << solve(T) << "\n";
    }
}
 
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

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