[斯特林數] Codeforces 932E. Team Work

Solution$Solution$

可能也算是裸題了咯。
求

∑i=1n(ni)ik$$\sum _{i=1}^n(\genfrac{}{}{0ex}{}{n}{i})i^k$$

用斯特林數把ik$i^k$ 展開。

∑d=0kd!S(k,d)(nd)2n−d$$\sum _{d=0}^{k}d!S(k,d)(\genfrac{}{}{0ex}{}{n}{d})2^{n-d}$$

O(k2)$\mathcal{O}(k^2)$ 直接做了啦~

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 5050;
const int MOD = 1000000007;

int n, k;
int S[N][N];
int fac[N], ifac[N], inv[N];

inline int pwr(int a, int b) {
    int c = 1;
    while (b) {
        if (b & 1) c = (ll)c * a % MOD;
        b >>= 1; a = (ll)a * a % MOD;
    }
    return c;
}
inline void pre(int n) {
    inv[1] = 1;
    for (int i = 2; i <= n; i++)
        inv[i] = (ll)(MOD - MOD / i) * inv[MOD % i] % MOD;
    fac[0] = ifac[0] = 1;
    for (int i = 1; i <= n; i++) {
        fac[i] = (ll)fac[i - 1] * i % MOD;
        ifac[i] = (ll)ifac[i - 1] * inv[i] % MOD;
    }
}
inline int C(int n, int m) {
    int res = 1;
    for (int i = 0; i < m; i++) {
        res = (ll)res * (n - i) % MOD;
        res = (ll)res * inv[i + 1] % MOD;
    }
    return res;
}

int main(void) {
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    cin >> n >> k;
    pre(k);
    S[0][0] = 1;
    for (int i = 1; i <= k; i++) {
        S[i][0] = 0;
        for (int j = 1; j <= i; j++)
            S[i][j] = ((ll)j * S[i - 1][j] + S[i - 1][j - 1]) % MOD;
    }
    int ans = 0;
    for (int d = 0; d <= k && d <= n; d++) {
        ans = (ans + (ll)fac[d] * S[k][d] % MOD * C(n, d) % MOD * pwr(2, n - d) % MOD) % MOD;
        // cerr << d << endl;
    }
    cout << ans << endl;
    return 0;
}