CF932ETeam Work

Posted 2022-10-06 shxnb666

题面

https://www.luogu.org/problem/CF932E

题解

先交换求和号、把$S2(i,j)$中$j>i$的部分去掉、再对角线替换、最后配个对$i$来说的“常数”，就把式子变成一个组合数了。

#include<iostream>
#include<cstdio>
using namespace std;
#define mod 1000000007
#define inv2 500000004
#define ri register int
#define N 5050
int pow(int a,int b)int s=1;while(b)if(b&1)s=1ll*s*a%mod;a=1ll*a*a%mod;b>>=1;return s;
int n,k,s2[N][N],ans;
int main() 
  scanf("%d%d",&n,&k);s2[0][0]=1;
  for(ri i=1;i<=k;++i)
    for(ri j=1;j<=k;++j)
      s2[i][j]=(s2[i-1][j-1]+1ll*s2[i-1][j]*j)%mod;
  for(ri j=0,pw=pow(2,n),nw=1;j<=min(n,k);pw=1ll*pw*inv2%mod,nw=1ll*nw*(n-j)%mod,++j)
    ans=(ans+1ll*s2[k][j]*nw%mod*pw%mod)%mod;
  printf("%d\n",ans);
  return 0;