CF1153F Serval and Bonus Problem
官方的解法是(O(n ^ 2))的，这里给出一个(O(n log n))的做法。
首先对于长度为(l)的线段，显然它的答案就是长度为(1)的线段的答案( imes l)，这样做只是为了方便计算。
考虑对于数轴上区间([0,1])内任意一个点(x)，它被一条随机线段覆盖的概率是多少：线段的两个端点都在它左边的概率是(x ^ 2)、都在它右边的概率是((1 - x) ^ 2)，那么它被覆盖的概率即为(p(x) = 1 - x^2 - (1 - x) ^ 2 = 2 x (1 - x))。
那么他被(ge k)条线段覆盖的概率为(f(x) = sum limits _{i = k} ^ n inom{n}{i} p(x) ^ i (1 - p(x)) ^ {n - i})。
根据定积分的定义就得到区间([0,1])内被(ge k)条线段覆盖的长度期望为(int _{0} ^{1} f(x) mathrm{d} x)。
现在我们的问题就是怎么算这个东西了：
[egin{align*} int _ 0 ^ 1 f(x) mathrm{d} x & = int _ 0 ^ 1 sum limits _{i = k} ^ n inom {n} {i} (2x (1 - x)) ^ i (1 - 2x (1 - x)) ^ {n - i} mathrm{d} x \ & = sum limits _{i = k} ^ n inom {n} {i} int _ 0 ^ 1 (2x (1 - x)) ^ i sum limits _{j = 0} ^ {n - i} inom {n - i} {j} (-2x(1 - x)) ^ j mathrm{d} x \ & = sum limits _{i = k} ^n inom {n} {i} sum limits _{j = 0} ^ {n - i} inom {n - i} {j} (-1) ^ j 2 ^ {(i + j)} int _ 0 ^ 1 x ^ {i + j} (1 - x) ^ {i + j} mathrm{d} x end{align*}
]
推到这里，就可以把积分去掉了，这是一个Beta Function的形式：记结论吧
[B(x, y) = int _ 0 ^ 1 t ^ {x - 1} (1 - t) ^ {y - 1} mathrm{d} t = frac {(x - 1) ! (y - 1) !} {(x + y - 1) !}
]
代入得：
[egin{align*} int _ 0 ^ 1 f(x) mathrm{d} x & = sum limits _{i = k} ^n inom {n} {i} sum limits _{j = 0} ^ {n - i} inom {n - i} {j} (-1) ^ j 2 ^ {(i + j)} frac {((i + j) ! ) ^ 2} {(2(i + j) + 1) !} \ & = n ! sum limits _{i = k} ^ n sum limits _ {j = 0} ^ {n - i} frac {1} {i !} frac {(-1) ^ j} {j !} frac {2 ^ {i + j} ((i + j) !) ^ 2} {(2(i+ j) + 1) ! (n - (i + j)) !} end{align*}
]
令(t = i + j)，(f[i] = frac {1} {i})，(g[j] = frac {(-1) ^ j} {j !})，(h[t] = frac{2^t (t !) ^ 2} {(2 t + 1) ! (n - t) !})，考虑枚举(t)：
[egin{align*} int _ 0 ^ 1 f(x) mathrm{d} x & = n ! sum limits _{t = k} ^ n h[t] sum limits _ {i = k} ^ t f[i] g[t - i] end{align*}
]
这后面就是一个非常显然的卷积式子了，直接FFT即可。
答案记得( imes l)。
//written by newbiechd
#include <cstdio>
#include <cctype>
#include <algorithm>
#define BUF 1000000
using namespace std;
const int N = 100003, yyb = 998244353, Gg = 3, Gi = 332748118;
char buf[BUF], *p1, *p2;
inline char gc() { return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, BUF, stdin), p1 == p2) ? EOF : *p1++; }
inline int rd() {
register int f = 0;
register char c;
while (!isdigit(c = gc())) {}
do
f = f * 10 + (c ^ 48);
while (isdigit(c = gc()));
return f;
}
int rev[N], G[2][N];
inline int power(int x, int y) {
register int o = 1;
for (; y; y >>= 1, x = 1ll * x * x % yyb)
if (y & 1)
o = 1ll * o * x % yyb;
return o;
}
inline void ntt(int *f, int len, int opt) {
register int i, j, k, x, y, p, q;
for (i = 1; i < len; ++i)
if (i < rev[i])
swap(f[i], f[rev[i]]);
for (i = 1; i < len; i <<= 1) {
p = G[opt][i];
for (j = 0; j < len; j += i << 1)
for (k = 0, q = 1; k < i; ++k, q = 1ll * p * q % yyb)
x = f[j | k], y = 1ll * q * f[i | j | k] % yyb, f[j | k] = (x + y) % yyb, f[i | j | k] = (x - y) % yyb;
}
}
int fac[N], invFac[N], f[N], g[N], h[N];
int main() {
#ifndef ONLINE_JUDGE
freopen("a.in", "r", stdin);
freopen("a.out", "w", stdout);
#endif
int n = rd(), m = n << 1 | 1, k = rd(), l = rd(), i, len, tmp, ans = 0;
fac[0] = 1;
for (i = 1; i <= m; ++i)
fac[i] = 1ll * fac[i - 1] * i % yyb;
invFac[m] = power(fac[m], yyb - 2);
for (i = m; i; --i)
invFac[i - 1] = 1ll * invFac[i] * i % yyb;
for (i = k; i <= n; ++i)
f[i] = invFac[i];
for (i = 0; i <= n; ++i)
g[i] = i & 1 ? yyb - invFac[i] : invFac[i];
for (len = 1; len <= m; len <<= 1) {}
for (i = 1; i < len; ++i)
rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? len >> 1 : 0);
for (i = 1; i < len; i <<= 1)
G[0][i] = power(Gg, (yyb - 1) / (i << 1)), G[1][i] = power(Gi, (yyb - 1) / (i << 1));
ntt(f, len, 0), ntt(g, len, 0);
for (i = 0; i < len; ++i)
f[i] = 1ll * f[i] * g[i] % yyb;
ntt(f, len, 1), tmp = power(len, yyb - 2);
for (i = 1; i <= n; ++i)
f[i] = 1ll * f[i] * tmp % yyb;
for (i = 1, tmp = 2; i <= n; tmp = (tmp << 1) % yyb, ++i)
h[i] = 1ll * fac[i] * fac[i] % yyb * tmp % yyb * invFac[i << 1 | 1] % yyb * invFac[n - i] % yyb;
for (i = k; i <= n; ++i)
ans = (1ll * f[i] * h[i] % yyb + ans) % yyb;
ans = 1ll * ans * fac[n] % yyb * l % yyb, printf("%d
", (ans + yyb) % yyb);
return 0;
}