CF932E Team Work

题目链接：洛谷

题目大意：求$$sum_{i=1}^nC_{n}^ii^k$$

数据范围：$1leq nleq 10^9,1leq kleq 5000$

这道题就是完全的模板了

$$ans=sum_{i=1}^nC_n^isum_{d=0}^kd!S(k,d)C_i^d$$

$$=sum_{d=0}^kd!S(k,d)sum_{i=1}^nC_n^iC_i^d$$

$$=sum_{d=0}^kd!S(k,d)C_n^dsum_{i=d}^nC_{n-d}^{i-d}$$

$$=sum_{d=0}^kd!S(k,d)C_n^d2^{n-d}$$

至于那个等式

$$C_n^iC_i^d=C_n^dC_{n-d}^{i-d}$$

只要根据组合数的定义拆一下就可以了。

#include<cstdio>
#define Rint register int
using namespace std;
typedef long long LL;
const int N = 5003, mod = 1e9 + 7;
int n, k, S[N][N], ans;
inline int kasumi(int a, int b){
    int res = 1;
    while(b){
        if(b & 1) res = (LL) res * a % mod;
        a = (LL) a * a % mod;
        b >>= 1;
    }
    return res;
}
int main(){
    scanf("%d%d", &n, &k);
    S[0][0] = 1;
    for(Rint i = 1;i <= k;i ++)
        for(Rint j = 1;j <= i;j ++)
            S[i][j] = ((LL) S[i - 1][j] * j + S[i - 1][j - 1]) % mod;
    int fac = 1, cho = 1;
    for(Rint i = 1;i <= k && i <= n;i ++){
        cho = (LL) cho * (n - i + 1) % mod * kasumi(i, mod - 2) % mod;
        fac = (LL) fac * i % mod;
        ans = (ans + (LL) fac * cho % mod * kasumi(2, n - i) % mod * S[k][i] % mod) % mod;
    }
    printf("%d", ans);
}