[题解] [CF932E] TeamWork

题面

题解

我们知道有
\[ n^k = \sum_{i = 0}^{n}i!\binom{n}{i}\begin{Bmatrix}k\\i\end{Bmatrix} \]
所以有
\[ \displaystyle\begin{aligned}\sum_{i = 1}^{n}\binom{n}{i}i^k&=\sum_{i=0}^{n}\binom{n}{i}i^k-\begin{bmatrix}k=0\end{bmatrix}\\&= \sum_{i = 0}^{n}\binom{n}{i}\sum_{j=0}^{i}j!\binom{i}{j}\begin{Bmatrix}k\\j\end{Bmatrix}-\begin{bmatrix}k=0\end{bmatrix}\\&=\sum_{i = 0}^{n}\sum_{j=0}^{i}j!\begin{Bmatrix}k\\j\end{Bmatrix}\binom{n}{i}\binom{i}{j}-\begin{bmatrix}k=0\end{bmatrix}\end{aligned} \]
交换求和号
\[ \displaystyle=\sum_{j=0}^{min(n, k)}j!\begin{Bmatrix}k\\j\end{Bmatrix}\sum_{i=j}^{n}\binom{n}{i}\binom{i}{j} \]
因为 \(i < j\) 时, \(\binom{i}{j} = 0\)
\[ \displaystyle=\sum_{j=0}^{min(n, k)}j!\begin{Bmatrix}k\\j\end{Bmatrix}\sum_{i=0}^{n}\binom{n}{i}\binom{i}{j}-\begin{bmatrix}k=0\end{bmatrix} \]
看到
\[ \displaystyle\sum_{i=0}^n\binom{n}{i}\binom{i}{j} \]
它的组合意义是, 从 \(n\) 个白色的球中选出 \(i\) 个数染成黑色, 再从 \(i\) 个黑球中选 \(j\) 个染成红色

因为有一个 \(\sum_{i=0}^n\) 所以最后的黑球个数可能是 \(\begin{bmatrix}0, n-j\end{bmatrix}\)

所以就相当于从 \(n\) 个白球中选 \(j\) 个染成红色, 其他的随便染不染都行

所以有
\[ \displaystyle\sum_{i = 0}^n\binom{n}{i}\binom{i}{j}=\binom{n}{j}*2^{n-j} \]
代入得
\[ \displaystyle=\sum_{j=0}^{min(n, k)}j!\begin{Bmatrix}k\\j\end{Bmatrix}\binom{n}{j}*2^{n-j}-\begin{bmatrix}k=0\end{bmatrix} \]
预处理斯特林数和组合数即可

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
const int N = 5005;
const int mod = 1e9 + 7; 
using namespace std;

int n, k, s[N][N], ans; 

template < typename T >
inline T read()
{
    T x = 0, w = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * w; 
}

int fpow(int x, int y)
{
    int res = 1;
    for( ; y; y >>= 1, x = 1ll * x * x % mod)
        if(y & 1) res = 1ll * res * x % mod;
    return res; 
}

int main()
{
    n = read <int> (), k = read <int> ();
    s[0][0] = 1;
    for(int i = 1; i <= k; i++)
        for(int j = 0; j <= i; j++)
            s[i][j] = (!j ? 0 : (s[i - 1][j - 1] + 1ll * j * s[i - 1][j] % mod) % mod);
    for(int res, j = 0; j <= k; j++)
    {
        if(j > n) break; 
        res = 1;
        for(int i = n; i > n - j; i--)
            res = 1ll * res * i % mod;
        ans = (ans + 1ll * s[k][j] * res % mod * fpow(2, n - j) % mod) % mod; 
    }
    if(!k) ans--; 
    printf("%d\n", ans); 
    return 0; 
}

来源：https://www.cnblogs.com/ztlztl/p/12204281.html