結果
問題 | No.430 文字列検索 |
ユーザー |
|
提出日時 | 2024-09-21 14:32:53 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 902 ms / 2,000 ms |
コード長 | 2,442 bytes |
コンパイル時間 | 2,923 ms |
コンパイル使用メモリ | 247,424 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-11-10 01:13:26 |
合計ジャッジ時間 | 11,958 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 14 |
ソースコード
#include<bits/stdc++.h>using namespace std;const uint64_t hashmod = 0x1fffffffffffffff;const uint64_t base = chrono::duration_cast<chrono::microseconds>(chrono::system_clock::now().time_since_epoch()).count() % hashmod;const uint64_t mdf = uint64_t(1e17) + 1;template<typename T>struct rolling_hash {int n;vector<uint64_t> hash, p;inline uint64_t mul(uint64_t a, uint64_t b) const {uint64_t a31 = a >> 31, b31 = b >> 31;a &= 0x7fffffffull; b &= 0x7fffffffull;uint64_t x = a * b31 + b * a31;uint64_t res = (a31 * b31 << 1) + (x >> 30) + ((x & 0x3fffffffull) << 31) + a * b;res = (res >> 61) + (res & hashmod);if (res >= hashmod) res -= hashmod;return res;}rolling_hash(const T &a) {n = (int)a.size();hash.assign(n + 1, 0ull);p.assign(n + 1, 0ull);p[0] = 1;for(int i = 0; i < n; i++) {p[i + 1] = mul(p[i], base);hash[i + 1] = mul(hash[i], base) + (mdf + (uint64_t)a[i]);if(hash[i + 1] >= hashmod) hash[i + 1] -= hashmod;}}// get hashing of S[l, r)inline uint64_t get(int l, int r) const {uint64_t res = hash[r] + hashmod - mul(hash[l], p[r - l]);if (res >= hashmod) res -= hashmod;return res;}// return longest common prefixes of a and binline int lcp(const rolling_hash &b, int l1, int r1, int l2, int r2) const {int size = min(r1 - l1, r2 - l2);int ok = 0, ng = size + 1;while(ng - ok > 1) {int mid = (ok + ng) / 2;if(get(l1, l1 + mid) == b.get(l2, l2 + mid)) ok = mid;else ng = mid;}return ok;}// return hash1 + hash2inline uint64_t connect(uint64_t hash1, uint64_t hash2, int hash2size) const {uint64_t res = mul(hash1, p[hash2size]) + hash2;if (res >= hashmod) res -= hashmod;return res;}};int main() {ios::sync_with_stdio(false);cin.tie(nullptr);string s;cin >> s;rolling_hash h(s);int m;cin >> m;long long ans = 0;for (int i = 0; i < m; i++) {string c;cin >> c;rolling_hash t(c);for (int j = 0; j + c.size() <= (int)s.size(); j++) {if (h.get(j, j+(c.size())) == t.get(0, (int) c.size())) ans++;}}cout << ans << "\n";}