BZOJ5093 图的价值——推式子+第二类斯特林数

原题链接

题解

题目等价于求这个式子

\[ans=n2^{\frac{(n-1)(n-2)}{2}}\sum\limits_{i=0}^{n-1}\binom{n-1}{i}i^k
\]

有这么一个式子

\[i^k=\sum\limits_{j=0}^{i}\begin{Bmatrix}
k\\
j
\end{Bmatrix}j!\binom{i}{j}\]

代入可得

\[ans=n2^{\frac{(n-1)(n-2)}{2}}\sum\limits_{i=0}^{n-1}\binom{n-1}{i}\sum\limits_{j=0}^{i}\begin{Bmatrix}
k\\
j
\end{Bmatrix}j!\binom{i}{j}\]

交换枚举顺序

\[ans=n2^{\frac{(n-1)(n-2)}{2}}\sum\limits_{j=0}^{n-1}\begin{Bmatrix}
k\\
j
\end{Bmatrix}j!\sum\limits_{i=j}^{n-1}\binom{n-1}{i}\binom{i}{j}\]

考虑到后面那个和号的组合意义为先在\(n-1\)个数中确定\(j\)个，剩下的可选可不选，即

\[ans=n2^{\frac{(n-1)(n-2)}{2}}\sum\limits_{j=0}^{n-1}\begin{Bmatrix}
k\\
j
\end{Bmatrix}j!\binom{n-1}{j}2^{n-1-j}
\]

\[=n2^{\frac{(n-1)(n-2)}{2}}\sum\limits_{j=0}^{n-1}\begin{Bmatrix}
k\\
j
\end{Bmatrix}\frac{(n-1)!}{(n-1-j)!}2^{n-1-j}\]

本题的\(n\)可能高达\(10^9\)，但是发现当\(j>k\)时\(\begin{Bmatrix}
k\\
j
\end{Bmatrix}\)为\(0\)，改一下求和上界

\[=n2^{\frac{(n-1)(n-2)}{2}}\sum\limits_{j=0}^{min\{n-1,k\}}\begin{Bmatrix}
k\\
j
\end{Bmatrix}\frac{(n-1)!}{(n-1-j)!}2^{n-1-j}\]

第二类斯特林数可以直接卷积出来，总复杂度\(O(nlogn)\)

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <ctime>
#include     <queue>
#include       <map>
#include       <set>

using namespace std;

#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define IINF 0x3f3f3f3f3f3f3f3fLL
#define DEF 0x8f8f8f8f
#define DDEF 0x8f8f8f8f8f8f8f8fLL
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back
#define re register
#define il inline

#define N 1000000
#define MOD 998244353

int n, k;
int a[N+5], b[N+5], S[N+5], fac[N+5], facinv[N+5];

int fpow(int x, int p) {
  int ret = 1;
  while(p) {
    if(p&1) ret = 1LL*ret*x%MOD;
    x = 1LL*x*x%MOD;
    p >>= 1;
  }
  return ret;
}

void bitReverse(int *s, int bit, int len) {
  static int tmp[4*N+5];
  tmp[0] = 0;
  for(int i = 1; i < len; ++i) {
    tmp[i] = (tmp[i>>1]>>1)|((i&1)<<(bit-1));
    if(i < tmp[i]) swap(s[i], s[tmp[i]]);
  }
}

void DFT(int *s, int bit, int len, int flag) {
  bitReverse(s, bit, len);
  for(int l = 1; l <= len; l <<= 1) {
    int mid = l>>1, t = fpow(3, (MOD-1)/l);
    if(flag) t = fpow(t, MOD-2);
    for(int *p = s; p != s+len; p += l) {
      int w = 1, x, y;
      for(int i = 0; i < mid; ++i) {
        x = p[i], y = 1LL*w*p[i+mid]%MOD;
        p[i] = (x+y)%MOD;
        p[i+mid] = (x-y)%MOD;
        w = 1LL*w*t%MOD;
      }
    }
  }
  if(flag) {
    int invlen = fpow(len, MOD-2);
    for(int i = 0; i < len; ++i) s[i] = 1LL*s[i]*invlen%MOD;
  }
}

int main() {
  scanf("%d%d", &n, &k);
  if(n == 1) {
    printf("0\n");
    return 0;
  }
  fac[0] = 1;
  for(int i = 1; i <= k; ++i) fac[i] = 1LL*fac[i-1]*i%MOD;
  facinv[k] = fpow(fac[k], MOD-2);
  for(int i = k; i >= 1; --i) facinv[i-1] = 1LL*facinv[i]*i%MOD;
  for(int i = 0; i <= k; ++i) {
    a[i] = facinv[i];
    if(i&1) a[i] *= -1;
    b[i] = 1LL*fpow(i, k)*facinv[i]%MOD;
  }
  int bit = 0, len;
  while((1<<bit) < 2*k+2) bit++;
  len = (1<<bit);
  DFT(a, bit, len, 0), DFT(b, bit, len, 0);
  for(int i = 0; i < len; ++i) S[i] = 1LL*a[i]*b[i]%MOD;
  DFT(S, bit, len, 1);
  int ans = 0, lim = min(n-1, k), x = 1, y = fpow(2, n-1), t = fpow(2, MOD-2);
  for(int i = 0; i <= lim; ++i) {
    ans = (ans+1LL*S[i]*x%MOD*y%MOD)%MOD;
    x = 1LL*x*(n-1-i)%MOD, y = 1LL*y*t%MOD;
  }
  if(n&1) ans = 1LL*n*fpow(fpow(2, (n-1)/2), n-2)%MOD*ans%MOD;
  else ans = 1LL*n*fpow(fpow(2, (n-2)/2), n-1)%MOD*ans%MOD;
  ans = (ans+MOD)%MOD;
  printf("%d\n", ans);
  return 0;
}