众所周知这题可以用斯特林数 (m^2) 求,但是要快速插值斯特林数等一堆东西估计过不了。
令 (f(x) = sumlimits_{i = 0}^{m} inom{x}{i} s_i)。(inom{x}{i}) 是 (i) 次多项式,因此存在这样的 (s) 。
二项式反演:(f(x) = sumlimits_{i = 0}^{x} inom{x}{i} s_i Leftrightarrow s_x = sumlimits_{i = 0}^{x} inom{x}{i} (-1)^{x - i} f(i))
通过卷积可以快速得到 (s) 数组。
通过这个我们可以快速得到一个点的点值。
(知道这个这题就不难了)
推式子:
[sumlimits_{k = 0}^{n} sumlimits_{i = 0}^{m}s_i inom{k}{i} inom{n}{k} x^k (1 - x)^{n - k}
]
[sumlimits_{k = 0}^{n} sumlimits_{i = 0}^{m}s_i inom{k}{i} inom{n}{k} x^k (1 - x)^{n - k}
]
[sumlimits_{k = 0}^{n} sumlimits_{i = 0}^{m} s_i inom{n}{i} inom{n - i}{k - i} x^k (1 - x)^{n - k}
]
[sumlimits_{i = 0}^{m} s_i inom{n}{i} sumlimits_{k = i}^{n} inom{n - i}{k - i} x^k (1 - x)^{n - k}
]
[sumlimits_{i = 0}^{m} s_i inom{n}{i} sumlimits_{k = 0}^{n - i} inom{n - i}{k} x^{k + i} (1 - x)^{n - k - i}
]
[sumlimits_{i = 0}^{m} s_i inom{n}{i} x^i sumlimits_{k = 0}^{n - i} inom{n - i}{k} x^{k} (1 - x)^{n - i - k}
]
[sumlimits_{i = 0}^{m} s_i inom{n}{i} x^i (x + 1 - x)^{n - i}
]
[sumlimits_{i = 0}^{m} s_i inom{n}{i} x^i
]
这样就可以很方便地做了。
时间复杂度 (Theta(m log m))
代码:
int fac[N], ifac[N];
void minit(int x) {
fac[0] = 1;
L(i, 1, x) fac[i] = (ll) fac[i - 1] * i % mod;
ifac[x] = qpow(fac[x]);
R(i, x, 1) ifac[i - 1] = (ll) ifac[i] * i % mod;
}
int fpow(int x) {
return x % 2 == 0 ? 1 : mod - 1;
}
int n, m, x, f[N], g[N], s[N], ans;
int main() {
n = read(), m = read(), x = read();
minit(m), init(m << 1);
L(i, 0, m) f[i] = (ll) ifac[i] * fpow(i) % mod;
L(i, 0, m) g[i] = (ll) ifac[i] * read() % mod;
Mul(f, g, s, m + 1, m + 1);
L(i, 0, m) s[i] = (ll) s[i] * fac[i] % mod;
int now = 1;
L(i, 0, m) (ans += (ll) now * s[i] % mod * ifac[i] % mod) %= mod, now = (ll) now * x % mod * (n - i) % mod;
cout << ans << endl;
return 0;
}
祝大家学习愉快!