2024-04-25 12:38:08 +01:00
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static vector<MLL<MDL1>> power1;
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static vector<MLL<MDL2>> power2;
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static const ll b = rd();
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template <typename _Tp>
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struct hash_vec {
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using hash_type = pll;
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MLL<MDL1> hash1;
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MLL<MDL2> hash2;
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vector<_Tp> seq;
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size_t size() {
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return seq.size();
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}
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void push_back(const _Tp& x) {
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hash1 = hash1 * b + x;
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hash2 = hash2 * b + x;
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seq.push_back(x);
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}
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void push_front(const _Tp& x) {
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size_t length = size();
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hash1 += x * power1[length];
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hash2 += x * power2[length];
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seq.push_front(x);
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}
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void pop_back() {
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_Tp e = seq.back(); seq.pop_back();
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hash1 = (hash1 - e) / b;
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hash2 = (hash2 - e) / b;
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}
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void pop_front() {
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_Tp e = seq.front(); seq.pop_front();
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int length = seq.size();
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hash1 -= e * power1[length];
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hash2 -= e * power2[length];
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}
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void set(size_t pos, const _Tp& value) {
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int length = seq.size();
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int old_value = seq[pos];
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hash1 += (value - old_value) * power1[length - 1 - pos];
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hash2 += (value - old_value) * power2[length - 1 - pos];
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seq[pos] = value;
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}
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const _Tp& operator[](size_t pos) {
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return seq[pos];
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}
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hash_type hash() {
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return {hash1.val, hash2.val};
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}
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void clear() {
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hash1 = 0;
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hash2 = 0;
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seq.clear();
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}
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hash_vec(size_t maxn) {
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clear();
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2024-05-27 15:52:42 +01:00
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MLL<MDL1> c1 = power1.size() ? power1.back() * b : 1;
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MLL<MDL2> c2 = power2.size() ? power2.back() * b : 1;
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2024-04-25 12:38:08 +01:00
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for (int i = power1.size(); i < maxn; ++i) {
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power1.push_back(c1);
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power2.push_back(c2);
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c1 *= b;
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c2 *= b;
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}
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}
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hash_vec(size_t maxn, const _Tp& init_value) : hash_vec(maxn) {
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for (size_t i = 0; i != maxn; ++i) {
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push_back(init_value);
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}
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}
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};
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2024-06-09 19:39:08 +01:00
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struct range_hash {
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vector<pair<MLL<MDL1>, MLL<MDL2>>> hp;
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template <typename T>
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range_hash(const T& vec) {
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hp.emplace_back();
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hash_vec<ll> hs(vec.size());
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for (auto&& x : vec) {
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hs.push_back(x);
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hp.emplace_back(hs.hash());
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}
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}
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2024-06-09 19:39:23 +01:00
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/// query hash of subarray [l, r]. Index starts from 0.
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2024-06-09 19:39:08 +01:00
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pair<MLL<MDL1>, MLL<MDL2>> range_query(size_t l, size_t r) {
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return {
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(hp[r + 1].first - hp[l].first * power1[r + 1 - l]),
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(hp[r + 1].second - hp[l].second * power2[r + 1 - l]),
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};
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}
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};
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