[BZOJ2839]集合计数

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

Sol

二项式反演

设$g_i$表示随机选若干个集合 交集至少为$i$个的方案
$g_i$=$C(n,i)$*($2^{2^{n-i}}$$-1$)
用$f_i$表示恰好i个的方案
则有$g_i$=Σ$C(i,j)*f_j$ $ j∈[1,n] $
然后就直接二项式反演
$f_i$=$ΣC(i,j)*g_j*(-1)^{j-i}$

#include <bits/stdc++.h>
using namespace std;
const long long fish=1e9+7;
const int maxn=1e6;
long long inv[maxn+5],fac[maxn+5];
long long dp[maxn+5];
long long Pow(long long x,int y){
    long long ans=1;
    for (;y;y>>=1){
        if (y%2) ans=ans*x%fish;
        x=x*x%fish;
    }
    return ans;
}
long long C(int N,int R){
    return fac[N]*1ll*inv[R]%fish*inv[N-R]%fish;
}
int main(){
    int N,K;
    scanf("%d%d",&N,&K);
    fac[0]=1;
    for (int i=1;i<=maxn;i++)
            fac[i]=fac[i-1]*1ll*i%fish;
    inv[maxn]=Pow(fac[maxn],fish-2);
    for (int i=maxn-1;i>=0;i--)
        inv[i]=inv[i+1]*1ll*(i+1)%fish;
    long long tmp=2;
    for (int i=N;i>=0;i--){
        dp[i]=C(N,i)*(tmp-1)%fish;
        tmp=tmp*tmp%fish;
    }
    long long ans=0;
    long long tmp1=0;
    for (int i=K,xs=1;i<=N;i++,xs*=(-1)){
        long long tmp1=C(i,K)*dp[i]%fish*xs;
        if (tmp1<0) (tmp1+=fish)%=fish;
        ans=(ans+tmp1)%fish;
    }
    cout<<ans;
}