sotanishy's code snippets for competitive programming
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#define PROBLEM "https://judge.yosupo.jp/problem/number_of_subsequences" #include "../../math/modint.hpp" #include "../../data-structure/fenwick_tree.hpp" #include <bits/stdc++.h> using namespace std; using mint = Modint<998244353>; struct AddMonoid { using T = mint; static T id() { return 0; } static T op(T a, T b) { return a + b; } }; int main() { ios_base::sync_with_stdio(false); cin.tie(0); int N; cin >> N; vector<int> A(N); for (auto& x : A) cin >> x; map<int, int> last; FenwickTree<AddMonoid> ft(N+1); ft.update(0, 1); for (int i = 0; i < N; ++i) { int l = last[A[i]]; ft.update(i+1, ft.prefix_fold(i+1) - ft.prefix_fold(l)); last[A[i]] = i + 1; } cout << ft.prefix_fold(N + 1) - 1 << endl; }
#line 1 "test/yosupo/number_of_subsequences.test.cpp" #define PROBLEM "https://judge.yosupo.jp/problem/number_of_subsequences" #line 2 "math/modint.hpp" #include <algorithm> #include <iostream> /** * @brief Mod int */ template <int m> class Modint { using mint = Modint; static_assert(m > 0, "Modulus must be positive"); public: static constexpr int mod() { return m; } constexpr Modint(long long y = 0) : x(y >= 0 ? y % m : (y % m + m) % m) {} constexpr int val() const { return x; } constexpr mint& operator+=(const mint& r) { if ((x += r.x) >= m) x -= m; return *this; } constexpr mint& operator-=(const mint& r) { if ((x += m - r.x) >= m) x -= m; return *this; } constexpr mint& operator*=(const mint& r) { x = static_cast<int>(1LL * x * r.x % m); return *this; } constexpr mint& operator/=(const mint& r) { return *this *= r.inv(); } constexpr bool operator==(const mint& r) const { return x == r.x; } constexpr mint operator+() const { return *this; } constexpr mint operator-() const { return mint(-x); } constexpr friend mint operator+(const mint& l, const mint& r) { return mint(l) += r; } constexpr friend mint operator-(const mint& l, const mint& r) { return mint(l) -= r; } constexpr friend mint operator*(const mint& l, const mint& r) { return mint(l) *= r; } constexpr friend mint operator/(const mint& l, const mint& r) { return mint(l) /= r; } constexpr mint inv() const { int a = x, b = m, u = 1, v = 0; while (b > 0) { int t = a / b; std::swap(a -= t * b, b); std::swap(u -= t * v, v); } return mint(u); } constexpr mint pow(long long n) const { mint ret(1), mul(x); while (n > 0) { if (n & 1) ret *= mul; mul *= mul; n >>= 1; } return ret; } friend std::ostream& operator<<(std::ostream& os, const mint& r) { return os << r.x; } friend std::istream& operator>>(std::istream& is, mint& r) { long long t; is >> t; r = mint(t); return is; } private: int x; }; #line 2 "data-structure/fenwick_tree.hpp" #include <functional> #include <vector> template <typename M> class FenwickTree { using T = M::T; public: FenwickTree() = default; explicit FenwickTree(int n) : n(n), data(n + 1, M::id()) {} T prefix_fold(int i) const { T ret = M::id(); for (; i > 0; i -= i & -i) ret = M::op(ret, data[i]); return ret; } void update(int i, const T& x) { for (++i; i <= n; i += i & -i) data[i] = M::op(data[i], x); } int lower_bound(const T& x) const { return lower_bound(x, std::less<>()); } template <typename Compare> int lower_bound(const T& x, Compare cmp) const { if (!cmp(M::id(), x)) return 0; int k = 1; while (k * 2 <= n) k <<= 1; int i = 0; T v = M::id(); for (; k > 0; k >>= 1) { if (i + k > n) continue; T nv = M::op(v, data[i + k]); if (cmp(nv, x)) { v = nv; i += k; } } return i + 1; } private: int n; std::vector<T> data; }; #line 5 "test/yosupo/number_of_subsequences.test.cpp" #include <bits/stdc++.h> using namespace std; using mint = Modint<998244353>; struct AddMonoid { using T = mint; static T id() { return 0; } static T op(T a, T b) { return a + b; } }; int main() { ios_base::sync_with_stdio(false); cin.tie(0); int N; cin >> N; vector<int> A(N); for (auto& x : A) cin >> x; map<int, int> last; FenwickTree<AddMonoid> ft(N+1); ft.update(0, 1); for (int i = 0; i < N; ++i) { int l = last[A[i]]; ft.update(i+1, ft.prefix_fold(i+1) - ft.prefix_fold(l)); last[A[i]] = i + 1; } cout << ft.prefix_fold(N + 1) - 1 << endl; }