[组合 斯特林数] Codeforces 932E. Team Work

ans=∑i=1n(ni)ik a n s = ∑ i = 1 n ( n i ) i k

用斯特林数展开 ik i k

ans=∑i=1n(ni)∑j=1kS(k,j)A(i,j) a n s = ∑ i = 1 n ( n i ) ∑ j = 1 k S ( k , j ) A ( i , j )

=∑j=1kS(k,j)∑i=1n(ni)A(i,j) = ∑ j = 1 k S ( k , j ) ∑ i = 1 n ( n i ) A ( i , j )

考虑 ∑ni=1(ni)A(i,j) ∑ i = 1 n ( n i ) A ( i , j ) 的组合意义

n n 个物品里选 j j 个的排列，剩下的 n−j n − j 可选可不选

那么 ∑ni=1(ni)A(i,j)=A(n,j)×2n−j ∑ i = 1 n ( n i ) A ( i , j ) = A ( n , j ) × 2 n − j

所以答案就是 ∑kj=1S(k,j)×A(n,j)×2n−j ∑ j = 1 k S ( k , j ) × A ( n , j ) × 2 n − j

O(k2) O ( k 2 )

#include 
#include 
#include 

using namespace std;

const int N=5010,P=1e9+7;

int S[N][N];

inline int Pow(int x,int y){
  int ret=1;
  for(;y;y>>=1,x=1LL*x*x%P) if(y&1) ret=1LL*ret*x%P;
  return ret;
}

inline int A(int n,int j){
  int ret=1;
  for(int i=0;i1LL*ret*(n-i)%P;
  return ret;
}

int main(){
  S[0][0]=1;
  for(int i=1;i<=5000;i++){
    for(int j=1;j<=i;j++)
      S[i][j]=(S[i-1][j-1]+1LL*j*S[i-1][j])%P;
  }
  int ans=0;
  int n,k; scanf("%d%d",&n,&k);
  for(int j=1;j<=k && j<=n;j++)
    ans=(ans+1LL*S[k][j]*A(n,j)%P*Pow(2,n-j))%P;
  printf("%d\n",ans);
  return 0;
}