結果

問題 No.2761 Substitute and Search
ユーザー GOTKAKO
提出日時 2025-06-10 16:14:02
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3,019 ms / 4,000 ms
コード長 8,830 bytes
コンパイル時間 2,741 ms
コンパイル使用メモリ 217,208 KB
実行使用メモリ 202,752 KB
最終ジャッジ日時 2025-06-10 16:14:25
合計ジャッジ時間 18,678 ms
ジャッジサーバーID
(参考情報)
judge5 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 13
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ソースコード

diff #

#include <bits/stdc++.h>
using namespace std;

template<int idx>
struct Montgomery{
    //2^62未満&奇数modのみ.
    //初めにsetmodする.
    using u64 = uint64_t;
    using u128 = __uint128_t;
 
    private:
    static u64 mod,N2,Rsq; //N*N2≡1(mod N);
    //Rsq = R^2modN; R=2^64.
    u64 v = 0;
    public:
    long long val(){return reduce(v);}
    static constexpr u64 getmod(){return mod;}
    static void setmod(u64 m){
        assert(m<(1LL<<62)&&(m&1));
        mod = m; N2 = mod;
        for(int i=0; i<5; i++) N2 *= 2-N2*mod;
        Rsq = (-u128(mod))%mod;
    }
    //reduce = T*R^-1modNを求める.
    u64 reduce(const u128 &T){
        //T*R^-1≡(T+(T*(-N2))modR*N)/R 2N未満なので-N必要かだけで良い.
        u64 ret = (T+u128(((u64)T)*(-N2))*mod)>>64;
        if(ret >= mod) ret -= mod;
        return ret;
    }
    //初期値<mod. 初めにw*R modN...->reduce(R^2)でok.
    Montgomery(){v = 0;} Montgomery(long long w):v(reduce(u128(w)*Rsq)){}
 
    Montgomery& operator=(const Montgomery &b) = default;
    Montgomery operator-()const{return Montgomery()-Montgomery(*this);}
    Montgomery operator+(const Montgomery &b)const{return Montgomery(*this)+=b;}
    Montgomery operator-(const Montgomery &b)const{return Montgomery(*this)-=b;}
    Montgomery operator*(const Montgomery &b)const{return Montgomery(*this)*=b;}
    Montgomery operator/(const Montgomery &b)const{return Montgomery(*this)/=b;}
    Montgomery& operator+=(const Montgomery &b){
        v += b.v;
        if(v >= mod) v -= mod;
        return (*this);
    }
    Montgomery& operator-=(const Montgomery &b){
        v += mod-b.v;
        if(v >= mod) v -= mod;
        return (*this);
    }
    Montgomery& operator*=(const Montgomery &b){
        v = reduce(u128(v)*b.v);
        return (*this);
    }
    Montgomery& operator/=(const Montgomery &b){
        (*this) *= b.inv();
        return (*this);
    }
    Montgomery pow(u64 b)const{
        Montgomery ret = 1,p = (*this);
        if(b < 0) p = p.inv(),b = -b;
        while(b){
            if(b&1) ret *= p;
            p *= p; b >>= 1;
        }
        return ret;
    }
    Montgomery inv()const{return pow(mod-2);}
 
    bool operator!=(const Montgomery &b)const{return v!=b.v;}
    bool operator==(const Montgomery &b)const{return v==b.v;}
};
template<int idx> 
uint64_t Montgomery<idx>::mod;
template<int idx> 
uint64_t Montgomery<idx>::N2;
template<int idx> 
uint64_t Montgomery<idx>::Rsq;
using mont = Montgomery<0>;

bool MillerRabin(long long N,const vector<long long> &A){
    mont::setmod(N);
    long long s = __builtin_ctzll(N-1),d = N-1;
    d >>= s;
    for(auto &a : A){
        if(N <= a) break;
        mont x = mont(a).pow(d);
        if(x != 1){
            long long t;
            for(t=0; t<s; t++){
                if(x == N-1) break;
                x *= x;
            }
            if(t == s) return false;
        }
    }
    return true;
}
bool isprime(const long long N){
    if(N <= 1) return false;
    else if(N == 2) return true;
    else if(N%2 == 0) return false;
    else if(N < 4759123141LL) return MillerRabin(N,{2,7,61});
    else return MillerRabin(N, {2,325,9375,28178,450775,9780504,1795265022});
}
random_device rnd;
mt19937 mt(rnd());
mt19937_64 mt2(rnd());
int getprime1(int minv,int range){
    assert(minv+range >= 2);
    int ret = -1;
    while(true){
        ret = mt()%range+minv;
        if(isprime(ret)) return ret;
    }
}
long long getprime2(long long minv,long long range){
    assert(minv+range >= 2);
    long long ret = -1;
    while(true){
        ret = mt2()%range+minv;
        if(isprime(ret)) return ret;
    }
}
 

template<typename mint> 
class SegmentTreeHash{
    private:
    using SS = tuple<int,mint,mint>;
    int siz = -1,n = -1;
    static bool First;
    static long long mod;
    static mint b;
    static vector<mint> pb;
    vector<SS> dat;
    
    void initMB(){
        if(First){
            First = false;
            mod = getprime2((long long)1e18,(long long)1e17);
            mint::setmod(mod);
            b = mt2()%1001001001+1001001001;
        }
        if(pb.size() < siz*2){
            pb.reserve(siz*2);
            if(pb.size() == 0) pb.emplace_back(mint(1));
            while(pb.size() < siz*2) pb.emplace_back(pb.back()*b);
        }
    }

    public:
    SS op(SS a, SS b){
        auto [len1,l1,r1] = a; //l 左から右,r 右から左.
        if(len1 == 0) return b;
        auto [len2,l2,r2] = b; 
        if(len2 == 0) return a;
        return {len1+len2,l1+l2*pb.at(len1),r1*pb.at(len2)+r2};
    } //ここpowにすれば連長圧縮なら長さ大きくてもいけそう.
    SS e(){return {0,0,0};}
    void renew (SS &a,SS x){
        a = op(a,x);
        //a = x; //set(pos,x)で可能.
        //その他.
    }    
    pair<long long,mint> getmodBase(){return {mod,b};}
    SegmentTreeHash(){}
    SegmentTreeHash(int N){init(N);}
    SegmentTreeHash(const vector<int> &A){//長さ配列サイズに合わせる.
        siz = 1; n = A.size();
        while(siz < n) siz *= 2;
        dat.resize(siz*2,e());
        initMB();
        for(int i=0; i<n; i++) dat.at(i+siz) = {1,A.at(i),A.at(i)}; //Baseの都合上intだけ?.
        for(int i=siz-1; i>0; i--) dat.at(i) = op(dat.at(i*2),dat.at(i*2+1));
    }
    SegmentTreeHash(const string &s){ 
        siz = 1; n = s.size();
        while(siz < n) siz *= 2;
        dat.resize(siz*2,e());
        initMB();
        for(int i=0; i<n; i++) dat.at(i+siz) = {1,s.at(i),s.at(i)}; //charでok.
        for(int i=siz-1; i>0; i--) dat.at(i) = op(dat.at(i*2),dat.at(i*2+1));
    }
    void init(int N){
        //全要素単位元に初期化.
        siz = 1; n = N;
        while(siz < n) siz *= 2;
        dat.assign(siz*2,e());
        initMB();
    }
    void init(const vector<int> &A){//長さ配列サイズに合わせる.
        siz = 1; n = A.size();
        while(siz < n) siz *= 2;
        dat.resize(siz*2,e());
        initMB();
        for(int i=0; i<n; i++) dat.at(i+siz) = {1,A.at(i),A.at(i)};
        for(int i=siz-1; i>0; i--) dat.at(i) = op(dat.at(i*2),dat.at(i*2+1));
    }
    void init(const string &s){
        siz = 1; n = s.size();
        while(siz < n) siz *= 2;
        dat.resize(siz*2,e());
        initMB();
        for(int i=0; i<n; i++) dat.at(i+siz) = {1,s.at(i),s.at(i)};
        for(int i=siz-1; i>0; i--) dat.at(i) = op(dat.at(i*2),dat.at(i*2+1));
    }
    void set(int pos,mint x){ //len=1限定.
        int memo = pos;
        pos = pos+siz;
        dat.at(pos) = {1,x,x};
        while(pos != 1){
            pos = pos/2;
            dat.at(pos) = op(dat.at(pos*2),dat.at(pos*2+1));
        }
    }
    void update(int pos,SS x){
        assert(false); //多分使わない.
        int memo = pos;
        pos = pos+siz;
        renew(dat.at(pos),x);
        while(pos != 1){
            pos = pos/2;
            dat.at(pos) = op(dat.at(pos*2),dat.at(pos*2+1));
        }
    }
    SS rangeans(int l, int r){
        SS retl = e(),retr = e();
        l += siz,r += siz;
        while(l < r){
            if(l&1) retl = op(retl,dat.at(l++));
            if(r&1) retr = op(dat.at(--r),retr);
            l >>= 1; r >>= 1;
        }
        return op(retl,retr);
    }
    SS get(int pos){return dat.at(pos+siz);}
    SS allrange(){return dat.at(1);}
    bool ispalindrome(int l,int r){ //[l,r)が等しいか 半開区間注意!
        auto [ign,hl,hr] = rangeans(l,r);
        return hl==hr;
    }
    mint gethash(const vector<int> &A){
        mint ret = 0,mul = 1;
        for(auto a : A) ret += mul*a,mul *= b;
        return ret;
    }
    mint gethash(const string &s){
        mont ret = 0,mul = 1;
        for(auto c : s) ret += mul*c,mul *= b;
        return ret;
    }
};
template<typename mint>
bool SegmentTreeHash<mint>::First=true;
template<typename mint>
long long SegmentTreeHash<mint>::mod;
template<typename mint>
mint SegmentTreeHash<mint>::b;
template<typename mint>
vector<mint> SegmentTreeHash<mint>::pb={};

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

    int N,L,Q; cin >> N >> L >> Q;
    vector<string> S(N);
    vector<SegmentTreeHash<mont>> Z(N);
    for(int i=0; i<N; i++){
        cin >> S.at(i);
        Z.at(i).init(S.at(i));
    }
    vector<mont> H(200);
    for(int i=1; i<200; i++) H.at(i) = i;
    while(Q--){
        int T; cin >> T;
        if(T == 1){
            int p; char c,d; cin >> p >> c >> d; p--;
            for(int i=0; i<N; i++) if(S.at(i).at(p) == c) S.at(i).at(p) = d,Z.at(i).set(p,H.at(d));
        }
        else{
            string t; cin >> t;
            mont now = Z.at(0).gethash(t);
            int answer = 0;
            for(int i=0; i<N; i++) answer += get<1>(Z.at(i).rangeans(0,t.size()))==now;
            cout << answer << "\n";
        }
    }
}
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