日常刷题2025-2-21
C. Beautiful Sequence子序列计数问题
https://codeforces.com/contest/2069/problem/C
思路：线性DP
子序列计数的问题是一类典型的dp，我回头专门讲一下。大概就是：求子序列的DP，一般对于某个元素可以区分：选/不选，来做。求两个子序列可以区分：不选/选进序列1/选进序列2，三种转移。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;#if !defined(ONLINE_JUDGE) && defined(LOCAL)
#include "helper.h"
#else
#define dbg(...) ;
#define local_go_m(x) int c;cin>>c;while(c--)x()
#endifconstexpr int M = 998244353;template<class T>
constexpr T power(T a, i64 b) {T res = 1;for (; b != 0; b /= 2, a *= a) {if (b & 1) {res *= a;}}return res;
}template<int M>
struct ModInt {
public:constexpr ModInt() : x(0) {}template<std::signed_integral T>constexpr ModInt(T x_) {T v = x_ % M;if (v < 0) {v += M;}x = v;}constexpr int val() const {return x;}constexpr ModInt operator-() const {ModInt res;res.x = (x == 0 ? 0 : M - x);return res;}constexpr ModInt inv() const {return power(*this, M - 2);}constexpr ModInt &operator*=(const ModInt &rhs) &{x = i64(x) * rhs.val() % M;return *this;}constexpr ModInt &operator+=(const ModInt &rhs) &{x += rhs.val();if (x >= M) {x -= M;}return *this;}constexpr ModInt &operator-=(const ModInt &rhs) &{x -= rhs.val();if (x < 0) {x += M;}return *this;}constexpr ModInt &operator/=(const ModInt &rhs) &{return *this *= rhs.inv();}friend constexpr ModInt operator*(ModInt lhs, const ModInt &rhs) {lhs *= rhs;return lhs;}friend constexpr ModInt operator+(ModInt lhs, const ModInt &rhs) {lhs += rhs;return lhs;}friend constexpr ModInt operator-(ModInt lhs, const ModInt &rhs) {lhs -= rhs;return lhs;}friend constexpr ModInt operator/(ModInt lhs, const ModInt &rhs) {lhs /= rhs;return lhs;}friend constexpr std::istream &operator>>(std::istream &is, ModInt &a) {i64 i;is >> i;a = i;return is;}friend constexpr std::ostream &operator<<(std::ostream &os, const ModInt &a) {return os << a.val();}friend constexpr std::strong_ordering operator<=>(ModInt lhs, ModInt rhs) {return lhs.val() <=> rhs.val();}private:int x;
};using Z = ModInt<M>;template<typename T>
struct Comb {std::vector<T> fac;std::vector<T> facInv;int n;Comb(int n_) {n = n_;fac.assign(n, T{});facInv.assign(n, T{});fac[0] = 1;for (int i = 1; i < n; i++) {fac[i] = fac[i - 1] * i;}for (int i = 0; i < n; i++) {facInv[i] = fac[i].inv();}}
};constexpr int COMB_N = 300300;
Comb<Z> comb(COMB_N);template<std::signed_integral T>
Z P(T n, T m) {assert(n >= 0 && m >= 0 && m <= n);if (n < comb.n) {return comb.fac[n] * comb.facInv[n - m];} else {T v = n, rnd = m;Z res = 1;while (rnd--) {res *= v--;}return res;}
}template<std::signed_integral T>
Z C(T n, T m) {return P(n, m) * comb.facInv[m];
}void go() {int n;cin >> n;vector<int> a(n);for (int i = 0; i < n; i++) cin >> a[i];vector<Z> f(n, 0);int c1 = 0;Z res = 0;for (int i = 0; i < n; i++) {if (a[i] == 1) {if (i) f[i] = f[i - 1];c1++;} else if (a[i] == 3) {if (i) f[i] = f[i - 1];res += f[i];} else {if (i) f[i] = f[i - 1] * 2;f[i] += c1;}}dbg(f);cout << res << endl;
}int main() {std::ios::sync_with_stdio(false);std::cin.tie(nullptr);local_go_m(go);return 0;
}
