제출 #533246

#제출 시각아이디문제언어결과실행 시간메모리
533246browntoadSnake Escaping (JOI18_snake_escaping)C++14
22 / 100
451 ms26260 KiB
#include <bits/stdc++.h>
#pragma GCC optimize ("Ofast", "unroll-loops")
using namespace std;
#define ll long long
//#define int ll
#define FOR(i,a,b) for (int i = (a); i<(b); i++)
#define REP(i,n) FOR(i,0,n)
#define REP1(i,n) FOR(i,1,n+1)
#define RREP(i,n) for (int i=(n)-1; i>=0; i--)
#define f first
#define s second
#define pb push_back
#define ALL(x) x.begin(),x.end()
#define SZ(x) (int)(x.size())
#define SQ(x) (x)*(x)
#define pii pair<int, int>
#define pdd pair<double ,double>
#define pcc pair<char, char> 
#define endl '\n'
//#define TOAD
#ifdef TOAD
#define bug(x) cerr<<__LINE__<<": "<<#x<<" is "<<x<<endl
#define IOS()
#else
#define bug(...)
#define IOS() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#endif

const ll inf = 1ll<<60;
const int iinf=2147483647;
const ll mod = 1e9+7;
const ll maxn=1594323;
const double PI=acos(-1);

ll pw(ll x, ll p, ll m=mod){
    ll ret=1;
    while (p>0){
        if (p&1){
            ret*=x;
            ret%=m;
        }
        x*=x;
        x%=m;
        p>>=1;
    }
    return ret;
}

ll inv(ll a, ll m=mod){
    return pw(a,m-2);
}

//=======================================================================================

int n,q;
string str;
int dp[maxn], pw3[15];
void SOS(){
	REP(i,(1<<n)){
		int ac=0;
		REP(j,n){
			if (i&(1<<j)) ac+=pw3[j];
		}
		dp[ac]=str[i]-'0';
	}
	REP(i,n){
		string cur;
		REP(j,n){
			cur+='0';
		}
		REP(j,pw3[n]){
			if (cur[i]=='2'){
				dp[j]+=dp[j-2*pw3[i]]+dp[j-pw3[i]];
			}
			REP(k,n){
				if (cur[k]!='2') {
					cur[k]++;
					break;
				}
				cur[k]='0';
			}
		}
	}
}
signed main (){
    IOS();
    cin>>n>>q;
    cin>>str;
    pw3[0]=1;
    REP1(i,14){
    	pw3[i]=pw3[i-1]*3;
    }
    SOS();
    string in;
    int cur=0;
    REP(i,q){
    	cur=0;
    	cin>>in;
    	reverse(ALL(in));
    	REP(i,n){
    		if (in[i]=='1'){
    			cur+=pw3[i];
    		}
    		else if (in[i]=='?'){
    			cur+=pw3[i]*2;
    		}
    	}
    	//cout<<cur<<endl;
    	cout<<dp[cur]<<endl;
    }
}
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