This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)
template<typename T>
void amin(T &a, T b) {
a = min(a,b);
}
template<typename T>
void amax(T &a, T b) {
a = max(a,b);
}
#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif
/*
*/
const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
void solve(int test_case)
{
int n,q; cin >> n >> q;
string s; cin >> s;
int B = max(n/2,1);
amin(B,7);
vector<int> pow3(n+5);
pow3[0] = 1;
rep1(i,n) pow3[i] = pow3[i-1]*3;
vector<pii> queries(q);
rep(id,q){
string t; cin >> t;
int first_t = 0, last_t = 0;
rep(i,sz(t)){
ll x = 0;
if(t[i] == '0') x = 0;
if(t[i] == '1') x = 1;
if(t[i] == '?') x = 2;
if(i < B) first_t = first_t*3+x;
else last_t = last_t*3+x;
}
queries[id] = {first_t,last_t};
}
vector<int> maskf(1<<(n-B));
rep(i,1<<(n-B)){
int mask = 0;
rep(bit,n-B){
int b = 0;
if(i&(1<<bit)) b = 1;
mask += pow3[bit]*b;
}
maskf[i] = mask;
}
vector<int> lsqf(pow3[n-B],inf1);
rep(i,pow3[n-B]){
if(i%3 == 2) lsqf[i] = 0;
else lsqf[i] = lsqf[i/3]+1;
}
int ok[1<<B][pow3[B]];
memset(ok,0,sizeof ok);
ok[0][0] = (1<<29)-1;
rep(mask1,1<<B){
rep(mask2,pow3[B]){
if(mask1 == 0 and mask2 == 0) conts;
int b1 = mask1&1;
int b2 = mask2%3;
if(b2 == 2 or b1 == b2){
ok[mask1][mask2] = ok[mask1>>1][mask2/3];
}
else{
ok[mask1][mask2] = 0;
}
}
}
vector<int> ans(q);
vector<int> dp(pow3[n-B]);
rep(mask1,1<<B){
fill(all(dp),0);
rep(i,1<<n){
int firstB = i>>(n-B);
int lastB = i^(firstB<<(n-B));
if(firstB != mask1) conts;
dp[maskf[lastB]] += s[i]-'0';
}
rep(mask2,pow3[n-B]){
int lsq = lsqf[mask2];
if(lsq >= inf1) conts;
rep(x,2){
int mask3 = mask2+pow3[lsq]*(x-2);
dp[mask2] += dp[mask3];
}
}
rep(id,q){
ans[id] += ok[mask1][queries[id].ff]&dp[queries[id].ss];
}
}
rep(i,q) cout << ans[i] << endl;
}
int main()
{
fastio;
int t = 1;
// cin >> t;
rep1(i, t) {
solve(i);
}
return 0;
}
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