Codeforces Round #551 (Div. 2) F. Serval and Bonus Problem
https://codeforces.com/contest/1153/problem/F
Without loss of generality, we assume \(L=1\). The probability that x is covered by a single interval is \(2x(1-x)\). Let \(f(x)=2x(1-x)\), then the expectation of the length covered by exactly i intervals can be written by
\[ \int_{0}^{1} \binom{N}{i} f(x)^i (1-f(x))^{N-i} dx \]
If we have
\[ \int_{0}^{1} f(x)^i dx \]
the answer can be computed easily. The coefficients of \(f(x)^i\) can be computed by \(f(x)^{i-1}\) in O(N) time. The total time complexity is O(N^2).
#include <iostream> #include <algorithm> #include <vector> #include <string> using namespace std; const int MOD = 998244353; struct mint { int n; mint(int n_ = 0) : n(n_) {} }; mint operator+(mint a, mint b) { a.n += b.n; if (a.n >= MOD) a.n -= MOD; return a; } mint operator-(mint a, mint b) { a.n -= b.n; if (a.n < 0) a.n += MOD; return a; } mint operator*(mint a, mint b) { return (long long)a.n * b.n % MOD; } mint &operator+=(mint &a, mint b) { return a = a + b; } mint &operator-=(mint &a, mint b) { return a = a - b; } mint &operator*=(mint &a, mint b) { return a = a * b; } mint modpow(mint a, int b) { mint res = 1; while (b > 0) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } mint modinv(mint a) { return modpow(a, MOD - 2); } mint F[100001] = {1, 1}; mint R[100001] = {1, 1}; mint I[100001] = {0, 1}; mint C(int n, int r) { if (n < 0 || r < 0 || n < r) return 0; return F[n] * R[n - r] * R[r]; } void init() { for (int i = 2; i <= 100000; i++) { I[i] = I[MOD % i] * (MOD - MOD / i); F[i] = F[i - 1] * i; R[i] = R[i - 1] * I[i]; } } mint dp[2020][4040]; mint idp[2020]; int main() { init(); int N, K, L; cin >> N >> K >> L; // P(x) = 2x(1-x) = -2x^2 + 2x dp[0][0] = 1; for (int i = 0; i < N; i++) { for (int j = 0; j <= 4000; j++) { dp[i + 1][j + 2] -= 2 * dp[i][j]; dp[i + 1][j + 1] += 2 * dp[i][j]; } } for (int i = 0; i <= N; i++) { for (int j = 0; j <= 4000; j++) { idp[i] += I[j + 1] * dp[i][j]; } } mint ans = 0; for (int i = K; i <= N; i++) { // \int_0^1 C(N,i) * P(x)^i * (1-P(x))^{N-i} for (int j = 0; j <= N - i; j++) { if (j % 2 == 0) { ans += C(N,i) * C(N-i,j) * idp[i + j]; } else { ans -= C(N,i) * C(N-i,j) * idp[i + j]; } } } ans *= L; cout << ans.n << endl; }