「CF932E」Team Work

Solution

题意：求\(\sum_{i=1}^n\begin{pmatrix}n\\i\end{pmatrix}\times i^k\)

大力颓柿子，根据常幂转下降幂公式，有

\[\sum_{i=1}^n\begin{pmatrix}n\\i\end{pmatrix}\times i^k\]

\[=\sum_{i=1}^n\begin{pmatrix}n\\i\end{pmatrix}\sum_{j=0}^i\begin{Bmatrix}k\\j\end{Bmatrix}\times i^{\underline j}\]

\[=\sum_{i=1}^n\begin{pmatrix}n\\i\end{pmatrix}\sum_{j=0}^i\begin{Bmatrix}k\\j\end{Bmatrix}\times j!\begin{pmatrix}i\\j\end{pmatrix}\]

\[=\sum_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\sum_{i=j}^n\frac{n!}{i!(n-i)!}\times j! \times \frac{i!}{j!(i-j)!}\]

\[=\sum_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\sum_{i=j}^n\frac{n!}{(n-i)!}\times \frac{1}{(i-j)!}\]

后面那个求和上下同乘\((n-j)!\)，有

\[\sum_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\sum_{i=j}^n\frac{n!}{(n-i)!}\times \frac{1}{(i-j)}\times\frac{(n-j)!}{(n-j)!}\]

\[=\sum_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\frac{n!}{(n-j)!}\sum_{i=j}^n\begin{pmatrix}n-j\\n-i\end{pmatrix}\]

\(i<j\)的时候组合数为\(0\)，不妨把\(i\)的下界变为\(0\)，于是有

\[\sum_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\frac{n!}{(n-j)!}\sum_{i=0}^n\begin{pmatrix}n-j\\n-i\end{pmatrix}\]

\[=\sum_{j=0}^{min(n,k)}\begin{Bmatrix}k\\j\end{Bmatrix}\frac{n!}{(n-j)!}\times 2^{n-j}\]

于是我们可以\(O(k^2)\)递推求出第二类斯特林数，枚举\(j\)时迭代计算\(\frac{n!}{(n-j)!}\)，快速幂计算\(2^{n-j}\)即可

Code

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #define inv(x) (fastpow((x),mod-2))using namespace std;typedef long long ll;template 
          
            void read (T &t){ t=0;int f=0;char c=getchar(); while(!isdigit(c)){f|=c=='-';c=getchar();} while(isdigit(c)){t=t*10+c-'0';c=getchar();} if(f)t=-t;}const int maxk=5000+5;const ll mod=1e9+7;ll n,k; ll s[maxk][maxk];ll fastpow(ll a,ll b){ ll re=1,base=a; while(b) { if(b&1) re=re*base%mod; base=base*base%mod; b>>=1; } return re;}int main(){ read(n),read(k); s[0][0]=1; for(register ll i=1;i<=k;++i) for(register ll j=1;j<=i;++j) s[i][j]=(s[i-1][j-1]+j*s[i-1][j]%mod)%mod; ll kkk=1,ans=0; for(register ll j=0;j<=min(n,k);++j) { ans=(ans+s[k][j]*kkk%mod*fastpow(2,n-j)%mod)%mod; kkk=kkk*(n-j)%mod; } printf("%lld",ans); return 0;}