結果

問題 No.2761 Substitute and Search
ユーザー 👑 binap
提出日時 2024-04-30 19:45:21
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
(最新)
AC  
(最初)
実行時間 -
コード長 4,216 bytes
コンパイル時間 4,729 ms
コンパイル使用メモリ 262,996 KB
最終ジャッジ日時 2025-02-21 09:55:43
ジャッジサーバーID
(参考情報)
judge1 / judge2
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 * 12 TLE * 1
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;
ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return
    os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : "");
    return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
random_device rd;
mt19937 mt(rd());
int random_prime(){
while (true){
int pr = mt() % 100000000 + 900000000;
if (atcoder::internal::is_prime_constexpr(pr)){
return pr;
}
}
}
using ull = unsigned long long;
using mint1 = atcoder::dynamic_modint<1>;
using mint2 = atcoder::dynamic_modint<2>;
struct mints{
mint1 d1;
mint2 d2;
mints(long long val = 0) : d1(val), d2(val) {}
mints(mint1 d1, mint2 d2) : d1(d1), d2(d2) {}
mints& operator++() {
d1++; d2++;
return *this;
}
mints& operator--() {
d1--; d2--;
return *this;
}
mints& operator+=(const mints& a) {
d1 += a.d1; d2 += a.d2;
return *this;
}
mints& operator-=(const mints& a) {
d1 -= a.d1; d2 -= a.d2;
return *this;
}
mints& operator*=(const mints& a) {
d1 *= a.d1; d2 *= a.d2;
return *this;
}
mints& operator/=(const mints& a) { return *this = *this * a.inv(); }
mints operator+() const { return *this; }
mints operator-() const { return mints(0) - *this; }
mints pow(long long n) const {
return mints(d1.pow(n), d2.pow(n));
}
mints inv() const {
return mints(d1.inv(), d2.inv());
}
mints operator+(const mints& a) const {
return mints(d1 + a.d1, d2 + a.d2);
}
mints operator-(const mints& a) const {
return mints(d1 - a.d1, d2 - a.d2);
}
mints operator*(const mints& a) const {
return mints(d1 * a.d1, d2 * a.d2);
}
mints operator/(const mints& a) const {
return mints(d1 / a.d1, d2 / a.d2);
}
bool operator==(const mints& a) const {
return d1 == a.d1 and d2 == a.d2;
}
bool operator!=(const mints& a) const {
return d1 != a.d1 or d2 != a.d2;
}
ull get_hash(){
return (ull(d1.val()) << 32) + ull(d2.val());
};
};
ostream& operator<<(ostream& os, const mints& a) {os << a.d1 << '&' << a.d2; return os;}
const int base = 334;
struct S{
mints val, c;
S(mints val, mints c) : val(val), c(c) {}
};
S op(S a, S b){
S res(a.val * b.c + b.val, a.c * b.c);
return res;
}
S e(){
S e(mints(0), mints(1));
return e;
}
int main(){
{
const int p1 = random_prime();
const int p2 = random_prime();
mint1::set_mod(p1);
mint2::set_mod(p2);
}
int n, l, q;
cin >> n >> l >> q;
vector<string> s(n);
cin >> s;
vector<segtree<S, op, e>> seg;
rep(i, n){
vector<S> init(l, S(mints(0), base));
rep(j, l) init[j] = S(mints(s[i][j] - 'a'), base);
segtree<S, op, e> seg_sub(init);
seg.push_back(seg_sub);
}
rep(_, q){
int t;
cin >> t;
if(t == 1){
int k;
char c, d;
cin >> k >> c >> d;
k--;
rep(i, n){
if(s[i][k] == c){
s[i][k] = d;
seg[i].set(k, S(mints(d - 'a'), base));
}
}
}
if(t == 2){
string t;
cin >> t;
int m = t.size();
mints rh(0);
int ans = 0;
rep(j, m){
rh *= base;
rh += t[j] - 'a';
}
rep(i, n){
auto [val, c] = seg[i].prod(0, m);
if(val == rh) ans++;
}
cout << ans << "\n";
}
}
return 0;
}
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