codeforces932E Team Work

题目链接：CF932E

由第二类斯特林数知

[n^m=sum_{i=0}^nS(m,i)*i!*dbinom{n}{i} ]

[egin{aligned} sum_{i=1}^n dbinom{n}{i}i^k&=sum_{i=1}^ndbinom{n}{i}sum_{j=0}^iS(k,j)*j!*dbinom{i}{j}\ &=sum_{i=1}^nfrac{n!}{i!(n-i)!}sum_{j=0}^iS(k,j)*j!*frac{i!}{(i-j)!j!}\ &=sum_{i=1}^nfrac{n!}{(n-i)!}sum_{j=0}^iS(k,j)*frac{1}{(i-j)!}\ &=sum_{j=0}^nS(k,j)sum_{i=j}^nfrac{n!}{(n-i)!(i-j)!}\ &=sum_{j=0}^nS(k,j)sum_{i=j}^nn^{underline{j}}dbinom{n-j}{n-i}\ &=sum_{j=0}^kS(k,j)n^{underline{j}}sum_{i=j}^ndbinom{n-j}{n-i}\ &=sum_{j=0}^kS(k,j)n^{underline{j}}2^{n-j} end{aligned} ]

(O(k^2))预处理第二类斯特林数，(O(k))计算答案，沿途维护(n^underline{j})和(2^{n-j})即可

#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<vector>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define lowbit(x) (x)&(-x)
#define rep(i,a,b) for (int i=a;i<=b;i++)
#define per(i,a,b) for (int i=a;i>=b;i--)
#define maxd 1000000007
#define inv2 500000004
typedef long long ll;
const double pi=acos(-1.0);
int n,k,N;
ll s[5050][5050];

int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

ll qpow(ll x,int y)
{
    ll ans=1;
    while (y)
    {
        if (y&1) ans=(ans*x)%maxd;
        x=(x*x)%maxd;
        y>>=1;
    }
    return ans;
}

int main()
{
    n=read();k=read();
    N=min(n,k);s[0][0]=1;
    rep(i,1,k)
    {
        rep(j,1,k) s[i][j]=(s[i-1][j-1]+s[i-1][j]*j)%maxd;
    }
    ll bin=qpow(2,n);
    ll ans=0,now=1;
    rep(i,0,N)
    {
        ans=(ans+s[k][i]*now%maxd*bin%maxd)%maxd;
        now=(now*(n-i))%maxd;bin=(bin*inv2)%maxd;
    }	
    printf("%lld",ans);
    return 0;
}