题解 | #F. 口吃#

题解发布于 CSND 😋

奉上 C++ 代码

#include <bits/stdc++.h>

using i64 = int64_t;
using u64 = uint64_t;
using f64 = double_t;
using i128 = __int128_t;

template<class T>
constexpr T power(T a, i64 b) {
    T res = 1;
    for (; b; b /= 2, a *= a) {
        if (b % 2) {
            res *= a;
        }
    }
    return res;
}
template<int P>
struct MInt {
    int x;
    constexpr MInt() : x{} {}
    constexpr MInt(i64 x) : x{norm(x % getMod())} {}
     
    static int Mod;
    constexpr static int getMod() {
        if (P > 0) {
            return P;
        } else {
            return Mod;
        }
    }
    constexpr static void setMod(int Mod_) {
        Mod = Mod_;
    }
    constexpr int norm(int x) const {
        if (x < 0) {
            x += getMod();
        }
        if (x >= getMod()) {
            x -= getMod();
        }
        return x;
    }
    constexpr int val() const {
        return x;
    }
    explicit constexpr operator int() const {
        return x;
    }
    constexpr MInt operator-() const {
        MInt res;
        res.x = norm(getMod() - x);
        return res;
    }
    constexpr MInt inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    constexpr MInt &operator*=(MInt rhs) & {
        x = 1LL * x * rhs.x % getMod();
        return *this;
    }
    constexpr MInt &operator+=(MInt rhs) & {
        x = norm(x + rhs.x);
        return *this;
    }
    constexpr MInt &operator-=(MInt rhs) & {
        x = norm(x - rhs.x);
        return *this;
    }
    constexpr MInt &operator/=(MInt rhs) & {
        return *this *= rhs.inv();
    }
    friend constexpr MInt operator*(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res *= rhs;
        return res;
    }
    friend constexpr MInt operator+(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res += rhs;
        return res;
    }
    friend constexpr MInt operator-(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res -= rhs;
        return res;
    }
    friend constexpr MInt operator/(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res /= rhs;
        return res;
    }
    friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
        i64 v;
        is >> v;
        a = MInt(v);
        return is;
    }
    friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }
    friend constexpr bool operator==(MInt lhs, MInt rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool operator!=(MInt lhs, MInt rhs) {
        return lhs.val() != rhs.val();
    }
};
 
template<>
int MInt<0>::Mod = 998244353;
 
template<int V, int P>
constexpr MInt<P> CInv = MInt<P>(V).inv();
 
constexpr int P = 1000000007;
using Z = MInt<P>;

// 设 E[i] 表示从 i 开始到 n 的期望
// 经整理是三对角矩阵，所以可以只记录对角线上的系数，以及右侧的（增广）

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    std::cout << std::fixed << std::setprecision(12);
    
    int n;
    std::cin >> n;

    std::vector<int> a(n - 1);
    for (int i = 0; i < n - 1; i++) {
        std::cin >> a[i];
    }
    std::vector<int> b(n - 1);
    for (int i = 0; i < n - 1; i++) {
        std::cin >> b[i];
    }

    std::vector<Z> x(n - 1);
    std::vector<Z> y(n - 1);
    std::vector<Z> z(n - 1);
    x[0] = Z(a[0]) / (a[0] + b[0]);
    y[0] = Z(b[0]) / (a[0] + b[0]) - 1;
    for (int i = 1; i < n - 1; i++) {
        Z dn = Z(a[i] + b[i]) * (a[i] + b[i]);
        x[i] = Z(a[i]) * a[i] / dn;
        y[i] = 2 * Z(a[i]) * b[i] / dn - 1;
        z[i] = Z(b[i]) * b[i] / dn;
    }

    std::vector<Z> s(n - 1, -1);
    for (int i = 0; i < n - 1; i++) {
        s[i] /= y[i];
        x[i] /= y[i];
        if (i + 1 < n - 1) {
            s[i + 1] -= z[i + 1] * s[i];
            y[i + 1] -= z[i + 1] * x[i];
        }
    }
    for (int i = n - 3; i >= 0; i--) {
        s[i] -= s[i + 1] * x[i];
    }

    std::cout << s[0] + 1 << "\n";

    return 0;
}