题面

Bzoj

Sol

一张无向无重边自环的图的边数最多为\(\frac{n(n-1)}{2}\)
考虑每个点的贡献
\[n*2^{\frac{n(n-1)}{2} - (n-1)}\sum_{i=0}^{n-1}i^kC(n-1, i)\]
很好理解
考虑后面的\(\sum_{i=0}^{n-1}i^kC(n-1, i)\)
\(i^k\)这里把它用第二类斯特林数表示出来
那么就是
\[\sum_{i=0}^{n-1}\sum_{j=0}^{i}S(k, j) j!C(i, j)\]
\[=\sum_{j=0}^{n-1}S(k, j)j!\sum_{i=j}^{n-1}C(n-1,i)C(i,j)\]

考虑\(\sum_{i=j}^{n-1}C(n-1,i)C(i,j)\)
就是\(C(n-1, j)\sum_{i=j}^{n-1}C(n-1, i-j)=C(n-1,j)2^{n-1-j}\)

带回去
\[\sum_{j=0}^{n-1}j!C(n-1,j)2^{n-1-j}S(k, j)\]
\[=\sum_{j=0}^{n-1}\frac{(n-1)!}{(n-1-j)!}2^{n-1-j}S(k,j)\]
又由于\(i>j\)时\(S(i, j)=0\)，\(n\)很大枚到\(k\)就可以了

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int Zsy(998244353);
const int Phi(998244352);
const int G(3);
const int _(8e5 + 5);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, k, ans, A[_], B[_], l, N, r[_], mx, fac[_], inv[_];

IL int Pow(RG ll x, RG ll y){
    RG ll ret = 1;
    for(; y; y >>= 1, x = x * x % Zsy) if(y & 1) ret = ret * x % Zsy;
    return ret;
}

IL void NTT(RG int* P, RG int opt){
    for(RG int i = 0; i < N; ++i) if(i < r[i]) swap(P[i], P[r[i]]);
    for(RG int i = 1; i < N; i <<= 1){
        RG int W = Pow(G, Phi / (i << 1));
        if(opt == -1) W = Pow(W, Zsy - 2);
        for(RG int p = i << 1, j = 0; j < N; j += p)
            for(RG int w = 1, k = 0; k < i; ++k, w = 1LL * w * W % Zsy){
                RG int X = P[k + j], Y = 1LL * w * P[k + j + i] % Zsy;
                P[k + j] = (X + Y) % Zsy, P[k + j + i] = (X - Y + Zsy) % Zsy;
            }
    }
    if(opt == 1) return;
    RG int Inv = Pow(N, Zsy - 2);
    for(RG int i = 0; i < N; ++i) P[i] = 1LL * P[i] * Inv % Zsy;
}

IL void Mul(){
    for(N = 1; N <= mx + mx; N <<= 1) ++l;
    for(RG int i = 0; i < N; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    NTT(A, 1); NTT(B, 1);
    for(RG int i = 0; i < N; ++i) A[i] = 1LL * A[i] * B[i] % Zsy;
    NTT(A, -1);
}

IL void Up(RG int &x, RG int y){
    x += y;
    if(x >= Zsy) x -= Zsy;
}

int main(RG int argc, RG char* argv[]){
    n = Input(), k = Input(), mx = min(n - 1, k);
    fac[0] = 1;
    for(RG int i = 1; i <= mx; ++i) fac[i] = 1LL * i * fac[i - 1] % Zsy;
    inv[mx] = Pow(fac[mx], Zsy - 2);
    for(RG int i = mx - 1; ~i; --i) inv[i] = 1LL * inv[i + 1] * (i + 1) % Zsy;
    for(RG int i = 0; i <= mx; ++i){
        A[i] = B[i] = inv[i];
        B[i] = 1LL * B[i] * Pow(i, k) % Zsy;
        if(i & 1) A[i] = Zsy - A[i];
    }
    Mul(); RG int Inv = Pow(2, Zsy - 2);
    for(RG int i = 0, e = 1, x = n - 1, pw = Pow(2, n - 1); i <= mx; ++i, --x){
        Up(ans, 1LL * e * pw % Zsy * A[i] % Zsy);
        e = 1LL * e * x % Zsy;
        pw = 1LL * pw * Inv % Zsy;
    }
    ans = 1LL * n * Pow(2, 1LL * n * (n - 1) / 2 - n + 1) % Zsy * ans % Zsy;
    printf("%d\n", ans);
    return 0;
}

原文地址：https://www.cnblogs.com/cjoieryl/p/8456959.html