[容斥原理] zoj 3556 How Many Sets I

主题链接：
http://acm.zju.edu.cn/onlinejudge/showProblem.do?
problemId=4535

How Many Sets I

Time Limit: 2 Seconds Memory Limit: 65536 KB

Give a set S, |S| = n, then how many ordered set group (S1, S2, ..., Sk) satisfies S1 ∩ S2 ∩ ... ∩ Sk =
∅. (Si is a subset of S, (1 <= i <= k))

Input

The input contains multiple cases, each case have 2 integers in one line represent n and k(1 <= k <= n <= 231-1), proceed to the end of
the file.

Output

Output the total number mod 1000000007.

Sample Input

1 1
2 2

Sample Output

1
9

Author: QU, Zhe

Contest: ZOJ Monthly, October 2011

Submit Status

题目意思：

已知|S|=n。给定k,求S1 ∩ S2 ∩
... ∩ Sk = ∅,当中Si是S的子集（i<=k)的种数。

n,k<=2^31-1

解题思路：

容斥原理

反向考虑。如果S1 ∩ S2 ∩
... ∩ Sk 不等于 ∅。则至少存在一个元素S1。S2。...，Sk都包括。

枚举都包括的元素.总的种数为(2^n)^k=2^(nk)

假设至少都包括一个元素，则种数为C(n,1)*(2^(n-1))^k=C(n,1)*2^((n-1)k)

假设至少都包括两个元素，则种数为C(n,2)*(2^(n-2))^k=C(n,2)*2^((n-2)k)

假设至少都包括i个元素，则种数为C(n,i)*(2^(n-i))^k=C(n,i)*2^((n-i)k)

减去包括一个的加上包括两个的减去包括3个的，如此类推。能够得出一下公式：

2^(nk)+C(n,1)*2^((n-1)k)-C(n,2)*2^((n-2)k)+...(-1)^i*C(n,i)*2^((n-i)k)+.....=(2^k-1)^n
(通过二项式公式）

所以答案转化为求(2^k-1)^n了，直接高速幂就可以。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

LL n,k;

LL quick(LL a,LL b)
{
LL res=1;

while(b)
{
if(b&1)
res=(res*a)%M;
b>>=1;
a=a*a%M;
}
return res;
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(~scanf("%lld%lld",&n,&k))
{
LL ans=(quick(2,k)-1+M)%M;

ans=quick(ans,n);

printf("%lld\n",ans);
}
return 0;
}