bzoj2839 集合计数

题面： 
一个有N个元素的集合有2^N个不同子集（包含空集），现在要在这2^N个集合中取出若干集合（至少一个），使得 
它们的交集的元素个数为K，求取法的方案数，答案模1000000007。

https://www.cnblogs.com/mrclr/p/10986797.html

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define ULL unsigned long long
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define dep(i,j,k) for(int i=k;i>=j;i--)
#define INF 0x3f3f3f3f
#define mem(i,j) memset(i,j,sizeof(i))
#define make(i,j) make_pair(i,j)
#define pb push_back
using namespace std;
const int mod = 1e9 + 7, N = 1000010;
LL n, k, ans;
LL ine[N], jc[N], jcc[N];
LL cal(LL x, LL y) {
    return jc[x] * jcc[y] % mod * jcc[x - y] % mod;
}
int main() {
    scanf("%lld %lld", &n, &k);
    ine[1] = jc[1] = jcc[1] = jc[0] = jcc[0] = 1;
    for(LL i = 2; i <= n; i++) {
        ine[i] = (mod - (mod / i) * ine[mod % i]) % mod;
        jcc[i] = jcc[i - 1] * ine[i] % mod;
        jc[i] = jc[i - 1] * i % mod;
    }
    for(LL i = n, flag = ((n - k) & 1) ? -1 : 1, tmp = 1; i >=k; i--) {
        ans = (ans + mod + flag * cal(i, k) * cal(n, i) % mod * tmp % mod) % mod;
        flag = -flag, tmp = tmp * (tmp + 2) % mod;
    }
    printf("%lld
", ans);
    return 0;
}