CF932E Team Work 题解

Description

给定 $n,k$，求：

$${\displaystyle \sum _{i=1}^n(\genfrac{}{}{0ex}{}{n}{i})\times i^k}$$

$1\le k\le 5000,1\le n\le {10}^9$。

Solution

看到那个 $i^k$ 很不爽，但是 $k$ 很小，考虑用斯特林数改写一下：

$$i^k=\sum _{j=0}^k(\genfrac{}{}{0ex}{}{i}{j})\left\{\begin{array}{c}k\\ j\end{array}\right\}\cdot j!$$

代回原式得：

$${\displaystyle \begin{array}{rl}& \sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{i})\cdot \sum _{j=0}^k(\genfrac{}{}{0ex}{}{i}{j})\left\{\begin{array}{c}k\\ j\end{array}\right\}j!\\ =& \sum _{j=0}^{k}j!\left\{\begin{array}{c}k\\ j\end{array}\right\}\cdot \sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{i})(\genfrac{}{}{0ex}{}{i}{j})\\ =& \sum _{j=0}^{k}j!\left\{\begin{array}{c}k\\ j\end{array}\right\}\cdot \sum _{i=j}^n\frac{n!}{i!(n-i)!}\cdot \frac{i!}{j!(i-j)!}\\ =& n!\sum _{j=0}^k\left\{\begin{array}{c}k\\ j\end{array}\right\}\cdot \sum _{i=j}^n\frac{(\genfrac{}{}{0ex}{}{n-j}{i-j})}{(n-j)!}\\ =& n!\sum _{j=0}^k\frac{1}{(n-j)!}\left\{\begin{array}{c}k\\ j\end{array}\right\}\sum _{i=0}^{n-j}(\genfrac{}{}{0ex}{}{n-j}{i})\\ =& n!\sum _{j=0}^k\frac{2^{n-j}}{(n-j)!}\cdot \left\{\begin{array}{c}k\\ j\end{array}\right\}\end{array}}$$

于是直接预处理出斯特林数即可做到 $O(k^2+k\log n)$，如果用卷积预处理的话就可以做到 $O(k\log k+k\log n)$。

Code

#include <bits/stdc++.h>
 
// #define int int64_t
 
using i64 = int64_t;
 
const int kMod = 1e9 + 7;
 
int s[5005][5005];
 
int qpow(int bs, int idx = kMod - 2) {
  int ret = 1;
  for (; idx; idx >>= 1, bs = (i64)bs * bs % kMod)
    if (idx & 1)
      ret = (i64)ret * bs % kMod;
  return ret;
}
 
void dickdreamer() {
  int n, k, ans = 0;
  std::cin >> n >> k;
  s[0][0] = 1;
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= i; ++j) {
      s[i][j] = (s[i - 1][j - 1] + (i64)j * s[i - 1][j] % kMod) % kMod;
    }
  }
  for (int i = 0, c = 1; i <= std::min(n, k); ++i) {
    ans = (ans + (i64)s[k][i] * c % kMod * qpow(2, n - i) % kMod) % kMod;
    c = (i64)c * (n - i) % kMod;
  }
  std::cout << ans << '\n';
}
 
int32_t main() {
#ifdef ORZXKR
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
#endif
  std::ios::sync_with_stdio(0), std::cin.tie(0), std::cout.tie(0);
  int T = 1;
  // std::cin >> T;
  while (T--) dickdreamer();
  // std::cerr << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
  return 0;
}