P4233 射命丸文的笔记

P4233 射命丸文的笔记

考虑所有竞赛图上所有哈密顿路径的数量。考虑圆排列的数量 $(n - 1)!$ 个，剩下的 $\frac {n (n - 1)} 2 - n$ 条边随便选择，那么答案为 $(n - 1)! 2 ^ {\frac {n (n - 1)} 2 - n}$ 。

考虑所有竞赛图中有哈密顿路径的数量。注意到图上缩点后的形态，一定是一个链状的图，每一个点向后面的点连边，如果有一个哈密顿路径，那么所有点都在一个强连通分量上。$n$ 个点的竞赛图有生成函数 $F(x) = \sum _ {i = 0} 2 ^ {\binom n 2} x ^ i$ 。考虑其中有多少个强连通块，将每个点分到其中一个强连通分量，强连通分量间有顺序，强连通分量内没有顺序，考虑 EGF ，$G(x)$ 表示 $n$ 个点的只有一个强连通分量的图的个数，那么 $k$ 个强连通分量即 $G(x) ^ k$ 。那么有 $F(x) = \sum _ {i = 0} G(x)$ ，即 $G(x) = 1 - \frac 1 {F(x)}$ 。

最终答案 $\frac {(n - 1)! 2 ^ {\frac {n (n - 1)} 2 - n}} {n!G(n)}$ 。

查看代码

#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        flag |= c == '-';
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
    if (x < 0) putchar('-'), x = ~x + 1;
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 5e5 + 10, mod = 998244353, inv2 = mod + 1 >> 1;
int rev[N];
int binpow (int b, int k = mod - 2)
{
    int res = 1;
    for (; k; k >>= 1, b = (LL)b * b % mod)
        if (k & 1) res = (LL)res * b % mod;
    return res;
}
void ntt (int *x, int bit, int op)
{
    int tot = 1 << bit;
    for (int i = 1; i < tot; ++i)
        if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
            swap(x[rev[i]], x[i]);
    for (int mid = 1; mid < tot; mid <<= 1)
    {
        int w1 = binpow(3, (mod - 1) / (mid << 1));
        if (!~op) w1 = binpow(w1);
        for (int i = 0; i < tot; i += mid << 1)
            for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
            {
                int p = x[i | j], q = (LL)k * x[i | j | mid] % mod;
                x[i | j] = (p + q) % mod, x[i | j | mid] = (p - q) % mod;
            }
    }
    if (~op) return;
    int itot = binpow(tot);
    for (int i = 0; i < tot; ++i)
        x[i] = (LL)x[i] * itot % mod;
}
void PolyInv(int n, int *x, int *g)
{
    if (n == 1) return void(g[0] = binpow(x[0]));
    int m = n + 1 >> 1;
    int bit = 0;
    while (1 << bit < n + m + m - 2) ++bit;
    int tot = 1 << bit;
    PolyInv(m, x, g);
    for (int i = m; i < tot; ++i) g[i] = 0;
    static int A[N];
    for (int i = 0; i < n; ++i) A[i] = x[i];
    for (int i = n; i < tot; ++i) A[i] = 0;
    ntt(g, bit, 1), ntt(A, bit, 1);
    for (int i = 0; i < tot; ++i)
        g[i] = (2 - (LL)g[i] * A[i]) % mod * g[i] % mod;
    ntt(g, bit, -1);
    for (int i = n; i < tot; ++i) g[i] = 0;
}
int main ()
{
    static int n, A[N], B[N];
    read(n);
    for (int i = 0, t = 1; i <= n; t = (LL)t * ++i % mod)
        A[i] = (LL)binpow(2, i * (i - 1ll) / 2 % (mod - 1)) * binpow(t) % mod;
    PolyInv(n + 1, A, B);
    for (int i = 0; i <= n; ++i) B[i] = -B[i];
    ++B[0];
    if (n >= 1) puts("1");
    if (n >= 2) puts("-1");
    for (int i = 3; i <= n; ++i, puts(""))
        write(B[i] ? ((LL)binpow(2, (i - 3ll) * i / 2 % (mod - 1)) * binpow((LL)B[i] * i % mod) % mod + mod) % mod: -1);
    return 0;
}