BZOJ 2839 集合计数

题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=2839

题解

考虑交集至少为$i$个的情况，设方案数为$g[i]$，显然
$g[i]=\binom{n}{i}(2^{2^{n-i}}-1)$
容斥一下，答案就是
$\sum_{i=k}^n (-1)^{i-k}\binom{i}{k}\binom{n}{i}(2^{2^{n-i}}-1)$

代码

#include <cstdio>

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

const int maxn=1000000;
const int mod=1000000007;

int n,k,fac[maxn+10],ifac[maxn+10],ans;

int quickpow(int a,int b,int m)
{
  int res=1;
  while(b)
    {
      if(b&1)
        {
          res=1ll*res*a%m;
        }
      a=1ll*a*a%m;
      b>>=1;
    }
  return res;
}

int C(int a,int b)
{
  return 1ll*fac[a]*ifac[b]%mod*ifac[a-b]%mod;
}

int main()
{
  n=read();
  k=read();
  fac[0]=1;
  for(int i=1; i<=n; ++i)
    {
      fac[i]=1ll*fac[i-1]*i%mod;
    }
  ifac[n]=quickpow(fac[n],mod-2,mod);
  for(int i=n-1; i>=0; --i)
    {
      ifac[i]=1ll*ifac[i+1]*(i+1)%mod;
    }
  int op=1;
  for(int i=k; i<=n; ++i)
    {
      ans=(ans+1ll*op*C(n-k,n-i)%mod*(quickpow(2,quickpow(2,n-i,mod-1),mod)-1+mod))%mod;
      op=mod-op;
    }
  printf("%lld\n",1ll*ans*C(n,k)%mod);
  return 0;
}