#include<bits/stdc++.h> #define ALL(X) X.begin(),X.end() #define FOR(I,A,B) for(int (I) = (A); (I) <= (B); (I)++) #define FORW(I,A,B) for(int (I) = (A); (I) < (B); (I)++) #define FORD(I,A,B) for(int (I) = (A); (I) >= (B); (I)--) #define CLEAR(X) memset(X,0,sizeof(X)) #define SIZE(X) int(X.size()) #define CONTAINS(A,X) (A.find(X) != A.end()) #define PB push_back #define MP make_pair #define X first #define Y second using namespace std; template<typename T, typename U> ostream& operator << (ostream& os, const pair<T, U> &_p) { return os << "(" << _p.X << "," << _p.Y << ")"; } template<typename T> ostream& operator << (ostream &os, const vector<T>& _V) { bool f = true; os << "["; for(auto v: _V) { os << (f ? "" : ",") << v; f = false; } return os << "]"; } template<typename T> ostream& operator << (ostream &os, const set<T>& _S) { bool f = true; os << "("; for(auto s: _S) { os << (f ? "" : ",") << s; f = false; } return os << ")"; } template<typename T, typename U> ostream& operator << (ostream &os, const map<T, U>& _M) { return os << set<pair<T, U>>(ALL(_M)); } typedef signed long long slong; typedef long double ldouble; const slong INF = 1000000100; const ldouble EPS = 1e-9; int N, M; string S; const int MAXN = 1000042; const slong p = 1000000021, mod = 1000000033, Q = 1000000007; slong pows[2*MAXN]; slong dp[MAXN]; typedef pair<int,int> pii; pii seg[MAXN]; // [st,en] klazul int nope[MAXN]; // nadklauzula, zignorowac vector<int> V[MAXN]; // 0 - x, 1 - ~x vector<slong> H[MAXN]; // hasze set<pair<int,int> > us; // (start,hasz) #define cerr if(0)cout void read_data() { scanf("%d ", &N); getline(cin, S); cerr << S << endl; vector<vector<int> > C; vector<pii> cur; vector<pair<pii,int> > w; // ((koniec,dlugosc),nr) FORW(i,0,SIZE(S)) { if(S[i] == 'x') { int neg = (S[i-1] == '~'); i++; int no = 0; while('0' <= S[i] && S[i] <= '9') { no = 10 * no + S[i] - '0'; i++; } cur.PB(MP(no,neg)); } if(S[i] == ')') { sort(ALL(cur)); w.PB(MP(MP(cur.back().X,cur.back().X-cur[0].X),SIZE(C))); C.PB(vector<int>()); for(pii x : cur) { C.back().PB(x.Y); } cur.clear(); } } sort(ALL(w)); M = SIZE(C); int cnt = 0; for(auto x : w) { cerr << x.X << " : " << C[x.Y] << endl; seg[cnt] = MP(x.X.X-x.X.Y,x.X.X); V[cnt] = C[x.Y]; cnt++; } } vector<pair<int,int> > starts[MAXN]; // (dlugosc,hasz) sie tam zaczynaja vector<int> kon[MAXN]; // ktore (nie nope) sie tam koncza slong bad[MAXN]; // ile prefiksow ze i jest pierwsza zla klauzula map<pair<int,int> ,int> meh[MAXN]; // sufiks (dlugosc,hasz) ma juz troche bad void solve() { FORW(i,0,M) { int n = SIZE(V[i]); H[i].resize(n,0); FORW(j,0,n) { if(j) { H[i][j] = H[i][j-1]; } if(V[i][j]) { H[i][j] = (H[i][j] + pows[seg[i].X+j])%mod; } } cerr << seg[i] << " : " << V[i] << endl; pair<int,slong> cur(seg[i].X,H[i].back()); if(us.count(cur)) { // duplikat klauzuli cerr << "duplikat" << endl; nope[i] = 1; continue; } us.insert(cur); // Sprawdz czy sa podklauzule FORW(j,0,n) { int jj = seg[i].X + j; if(nope[i]) { break; } for(auto y : starts[jj]) { if(y.X == n) { // ta sama dlugosc, duplikaty juz sprawdzone // to powinno dac O(n^(3/2)) bo porownujemy sie // tylko z krotszymi break; } int k = jj + y.X - 1; slong h = H[i][k - seg[i].X]; // nasz hasz na [jj,k] if(j) { h = (h + mod - H[i][j-1])%mod; } if(h == y.Y) { cerr << i << " ma podklazule w zakresie " << MP(jj,k) << endl; nope[i] = 1; break; } } } starts[seg[i].X].PB(MP(n,H[i].back())); if(!nope[i]) { kon[seg[i].Y].PB(i); } } dp[0] = 1; FOR(i,1,N) { dp[i] = (dp[i-1] * 2) % Q; for(int x : kon[i]) { bad[x] = dp[seg[x].X-1]; int n = seg[x].Y - seg[x].X + 1; FORW(j,0,n) { // ehh, to chyba tez wykryje duplikaty... pii ehh(j,H[x][j]); bad[x] -= meh[seg[x].X][ehh]; } bad[x] = ((bad[x]%Q)+Q)%Q; dp[i] -= bad[x]; FORW(j,0,n) { pii cur(n-j-1, H[x].back()); if(j) { cur.Y = (cur.Y + mod - H[x][j-1])%mod; } meh[seg[x].X+j][cur] += bad[x]; meh[seg[x].X+j][cur] %= Q; } } dp[i] = ((dp[i]%Q)+Q)%Q; } cout << dp[N] << "\n"; } int main() { pows[0] = 1; FORW(i,1,2*MAXN) { pows[i] = (pows[i-1] * p) % mod; } read_data(); solve(); }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 | #include<bits/stdc++.h> #define ALL(X) X.begin(),X.end() #define FOR(I,A,B) for(int (I) = (A); (I) <= (B); (I)++) #define FORW(I,A,B) for(int (I) = (A); (I) < (B); (I)++) #define FORD(I,A,B) for(int (I) = (A); (I) >= (B); (I)--) #define CLEAR(X) memset(X,0,sizeof(X)) #define SIZE(X) int(X.size()) #define CONTAINS(A,X) (A.find(X) != A.end()) #define PB push_back #define MP make_pair #define X first #define Y second using namespace std; template<typename T, typename U> ostream& operator << (ostream& os, const pair<T, U> &_p) { return os << "(" << _p.X << "," << _p.Y << ")"; } template<typename T> ostream& operator << (ostream &os, const vector<T>& _V) { bool f = true; os << "["; for(auto v: _V) { os << (f ? "" : ",") << v; f = false; } return os << "]"; } template<typename T> ostream& operator << (ostream &os, const set<T>& _S) { bool f = true; os << "("; for(auto s: _S) { os << (f ? "" : ",") << s; f = false; } return os << ")"; } template<typename T, typename U> ostream& operator << (ostream &os, const map<T, U>& _M) { return os << set<pair<T, U>>(ALL(_M)); } typedef signed long long slong; typedef long double ldouble; const slong INF = 1000000100; const ldouble EPS = 1e-9; int N, M; string S; const int MAXN = 1000042; const slong p = 1000000021, mod = 1000000033, Q = 1000000007; slong pows[2*MAXN]; slong dp[MAXN]; typedef pair<int,int> pii; pii seg[MAXN]; // [st,en] klazul int nope[MAXN]; // nadklauzula, zignorowac vector<int> V[MAXN]; // 0 - x, 1 - ~x vector<slong> H[MAXN]; // hasze set<pair<int,int> > us; // (start,hasz) #define cerr if(0)cout void read_data() { scanf("%d ", &N); getline(cin, S); cerr << S << endl; vector<vector<int> > C; vector<pii> cur; vector<pair<pii,int> > w; // ((koniec,dlugosc),nr) FORW(i,0,SIZE(S)) { if(S[i] == 'x') { int neg = (S[i-1] == '~'); i++; int no = 0; while('0' <= S[i] && S[i] <= '9') { no = 10 * no + S[i] - '0'; i++; } cur.PB(MP(no,neg)); } if(S[i] == ')') { sort(ALL(cur)); w.PB(MP(MP(cur.back().X,cur.back().X-cur[0].X),SIZE(C))); C.PB(vector<int>()); for(pii x : cur) { C.back().PB(x.Y); } cur.clear(); } } sort(ALL(w)); M = SIZE(C); int cnt = 0; for(auto x : w) { cerr << x.X << " : " << C[x.Y] << endl; seg[cnt] = MP(x.X.X-x.X.Y,x.X.X); V[cnt] = C[x.Y]; cnt++; } } vector<pair<int,int> > starts[MAXN]; // (dlugosc,hasz) sie tam zaczynaja vector<int> kon[MAXN]; // ktore (nie nope) sie tam koncza slong bad[MAXN]; // ile prefiksow ze i jest pierwsza zla klauzula map<pair<int,int> ,int> meh[MAXN]; // sufiks (dlugosc,hasz) ma juz troche bad void solve() { FORW(i,0,M) { int n = SIZE(V[i]); H[i].resize(n,0); FORW(j,0,n) { if(j) { H[i][j] = H[i][j-1]; } if(V[i][j]) { H[i][j] = (H[i][j] + pows[seg[i].X+j])%mod; } } cerr << seg[i] << " : " << V[i] << endl; pair<int,slong> cur(seg[i].X,H[i].back()); if(us.count(cur)) { // duplikat klauzuli cerr << "duplikat" << endl; nope[i] = 1; continue; } us.insert(cur); // Sprawdz czy sa podklauzule FORW(j,0,n) { int jj = seg[i].X + j; if(nope[i]) { break; } for(auto y : starts[jj]) { if(y.X == n) { // ta sama dlugosc, duplikaty juz sprawdzone // to powinno dac O(n^(3/2)) bo porownujemy sie // tylko z krotszymi break; } int k = jj + y.X - 1; slong h = H[i][k - seg[i].X]; // nasz hasz na [jj,k] if(j) { h = (h + mod - H[i][j-1])%mod; } if(h == y.Y) { cerr << i << " ma podklazule w zakresie " << MP(jj,k) << endl; nope[i] = 1; break; } } } starts[seg[i].X].PB(MP(n,H[i].back())); if(!nope[i]) { kon[seg[i].Y].PB(i); } } dp[0] = 1; FOR(i,1,N) { dp[i] = (dp[i-1] * 2) % Q; for(int x : kon[i]) { bad[x] = dp[seg[x].X-1]; int n = seg[x].Y - seg[x].X + 1; FORW(j,0,n) { // ehh, to chyba tez wykryje duplikaty... pii ehh(j,H[x][j]); bad[x] -= meh[seg[x].X][ehh]; } bad[x] = ((bad[x]%Q)+Q)%Q; dp[i] -= bad[x]; FORW(j,0,n) { pii cur(n-j-1, H[x].back()); if(j) { cur.Y = (cur.Y + mod - H[x][j-1])%mod; } meh[seg[x].X+j][cur] += bad[x]; meh[seg[x].X+j][cur] %= Q; } } dp[i] = ((dp[i]%Q)+Q)%Q; } cout << dp[N] << "\n"; } int main() { pows[0] = 1; FORW(i,1,2*MAXN) { pows[i] = (pows[i-1] * p) % mod; } read_data(); solve(); } |