CF932E Team Work

考虑将幂的形式转化为下降幂

$i^{k} = \sum_{j = 0}^{k} {\binom{k}{j}} i^{\underset{―}{j}}$

那么

$\begin{aligned} A N S & = \sum_{i = 1}^{n} (\binom{n}{i}) i^{k} \\ = \sum_{i = 1}^{n} (\binom{n}{i}) \sum_{j = 0}^{k} {\binom{k}{j}} i^{\underset{―}{j}} \\ = \sum_{j = 0}^{k} {\binom{k}{j}} j! \sum_{i = 0}^{n} (\binom{n}{i}) (\binom{i}{j}) \\ = \sum_{j = 0}^{k} {\binom{k}{j}} j! \sum_{i = 0}^{n} (\binom{n}{j}) (\binom{n - j}{i - j}) \\ = \sum_{j = 0}^{k} {\binom{k}{j}} j! (\binom{n}{j}) \sum_{i = 0}^{n} (\binom{n - j}{i - j}) \\ = \sum_{j = 0}^{k} {\binom{k}{j}} n^{\underset{―}{j}} \sum_{i = 0}^{n - j} (\binom{n - j}{i}) \\ = \sum_{j = 0}^{k} {\binom{k}{j}} n^{\underset{―}{j}} 2^{n - j} \end{aligned}$

其中 $(\binom{n}{j})$ 可以随着 $j$ 变化中维护。 $O (m^{2})$ ，算出第二类斯特林数。

查看代码

#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read(Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
    if (x < 0) putchar('-'), x = ~x + 1;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
typedef long long LL;
const int N = 5e3 + 10, mod = 1e9 + 7;
int n, m, s[N][N];
int binpow (int b, int k = mod - 2)
{
    int res = 1;
    for (; k; k >>= 1, b = (LL)b * b % mod)
        if (k & 1) res = (LL)res * b % mod;
    return res;
}
void init()
{
    s[0][0] = 1;
    for (int i = 1; i <= m; ++i)
        for (int j = 1; j <= m; ++j)
            s[i][j] = (s[i - 1][j - 1] + (LL)j * s[i - 1][j]) % mod;
}
int main ()
{
    read(n, m);
    init();
    int res = 0;
    for (int i = 1, t = n; i <= min(n, m); t = (LL)t * (n - i++) % mod)
        res = (res + (LL)t * s[m][i] % mod * binpow(2, n - i)) % mod;
    write(res);
    return 0;
}