[BZOJ 5093][Lydsy1711月赛]图的价值

danihao123 posted @ 2018年4月11日 14:53 in 题解 with tags bzoj 第二类斯特林数 NTT 斯特林反演 , 10 阅读
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统计的时候,考虑每个店对每个图的贡献,可以看出答案是:

\[n2^{\tfrac{(n - 1)(n - 2)}{2}}\,\sum_{i = 0}^{n - 1} \binom{n - 1}{i} i^k\]

求和号前面还好说,主要看求和号后面的咋搞。

后面的那一部分是个经典题吧……但我还是来推导一番吧:

\[
\begin{aligned}
\sum_{i = 0}^{m} \binom{m}{i} i^k&= \sum_{i = 0}^{m} \binom{m}{i}\sum_{j = 0}^k S(k, j) j!\binom{i}{j}\\
&= \sum_{j = 0}^k S(k, j) j! \sum_{i = 0}^m \binom{m}{i}\binom{i}{j} \\
&= \sum_{j = 0}^k S(k, j) m^{\underline{j}} \sum_{i = 0}^{m} \binom{n - j}{i - j} \\
&= \sum_{j = 0}^k S(k, j) m^{\underline{j}} 2^{m - j}
\end{aligned}
\]

然后如果我们能高效的求出某一行的第二类斯特林数,那么问题就迎刃而解了。

考虑斯特林反演(这本质上是一个容斥,用二项式反演也可以推导):

\[S(k, j) = \sum_{i = 0}^j \frac{(-1)^{j - i}}{(j - i)!} \cdot \frac{i^k}{i!}\]

这个东西很显然是一个卷积……而且模数还很友好(UOJ模数),用NTT求出一行的第二类斯特林数即可。

代码:

/**************************************************************
    Problem: 5093
    User: danihao123
    Language: C++
    Result: Accepted
    Time:22632 ms
    Memory:25824 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <algorithm>
#include <utility>
typedef long long ll;
const ll ha = 998244353LL;
const ll bs = 3LL;
inline ll pow_mod(const ll &a, ll b) {
  ll ans = 1LL, res = a;
  while(b) {
    if(1LL & b) ans = (ans * res) % ha;
    res = (res * res) % ha;
    b >>= 1;
  }
  return ans;
}
inline ll inv(const ll &x) {
  return pow_mod(x, ha - 2LL);
}
 
inline int flip(const int &bi, const int &x) {
  int ans = 0;
  for(int i = 0; i < bi; i ++) {
    if((1 << i) & x) {
      ans |= (1 << (bi - i - 1));
    }
  }
  return ans;
}
inline void ntt(ll *A, const int &bi, const int &len, bool flag = false) {
  for(int i = 0; i < len; i ++) {
    int v = flip(bi, i);
    if(v < i) std::swap(A[v], A[i]);
  }
  for(int L = 1; L < len; L <<= 1) {
    ll xi = pow_mod(bs, (ha - 1LL) / (ll(L << 1)));
    if(flag) xi = inv(xi);
    for(int i = 0; i < len; i += (L << 1)) {
      ll w = 1LL;
      for(int j = i; j < i + L; j ++) {
        ll x = A[j], y = A[j + L];
        A[j] = (x + (w * y) % ha) % ha;
        A[j + L] = (x - (w * y) % ha + ha) % ha;
        w = (w * xi) % ha;
      }
    }
  }
}
 
const int maxn = 800005;
ll A[maxn], B[maxn];
ll fac[maxn], down[maxn];
int main() {
  ll n; int k; scanf("%lld%d", &n, &k);
  if(n == 1) {
    puts("0"); return 0;
  }
  int len = 1, bi = 0;
  while(len <= (2 * k)) {
    len <<= 1; bi ++;
  }
  fac[0] = 1LL;
  for(int i = 1; i <= k; i ++) {
    fac[i] = (fac[i - 1] * (ll(i))) % ha;
  }
  for(int i = 0; i <= k; i ++) {
    A[i] = (inv(fac[i]) * pow_mod(ha - 1LL, i)) % ha;
  }
  for(int i = 0; i <= k; i ++) {
    B[i] = inv(fac[i]);
    B[i] = (B[i] * pow_mod(i, k)) % ha;
  }
  ntt(A, bi, len); ntt(B, bi, len);
  for(int i = 0; i < len; i ++) {
    A[i] = (A[i] * B[i]) % ha;
  }
  ntt(A, bi, len, true);
  ll inv_l = inv(len);
  for(int i = 0; i < len; i ++) {
    A[i] = (A[i] * inv_l) % ha;
  }
#ifdef LOCAL
  for(int i = 0; i <= k; i ++) {
    printf("%lld ", A[i]);
  }
  puts("");
#endif
  down[0] = 1LL;
  for(int i = 1; i <= std::min(ll(k), n - 1LL); i ++) {
    down[i] = (down[i - 1] * (n - (ll(i)))) % ha;
  }
  ll ans = 0LL;
  for(int i = 1; i <= std::min(ll(k), n - 1LL); i ++) {
    ll delta = (A[i] * down[i]) % ha;
    delta = (delta * pow_mod(2, n - 1LL - (ll(i)))) % ha;
    ans = (ans + delta) % ha;
  }
  ans = (ans * pow_mod(2LL, (n - 1LL) * (n - 2LL) / 2LL)) % ha;
  ans = (ans * n) % ha;
  printf("%lld\n", ans);
  return 0;
}

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