[容斥原理] zoj 3556 How Many Sets I

题目链接：

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

How Many Sets I

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Give a set S, |S| = n, then how many ordered set group (S₁, S₂, ..., S_k) satisfies S₁ ∩ S₂ ∩ ... ∩ S_k = ∅. (S_i 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 12 2

Sample Output

19

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Author: QU, Zhe
Contest: ZOJ Monthly, October 2011
Submit    Status

题目意思：

已知|S|=n，给定k,求S₁ ∩ S₂ ∩ ... ∩ S_k = ∅,其中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，假设S₁ ∩ S₂ ∩ ... ∩ S_k 不等于 ∅，则至少存在一个元素S1，S2，...，Sk都包含。

枚举都包含的元素.总的种数为(2^n)^k=2^(nk)

如果至少都包含一个元素，则种数为C(n,1)*(2^(n-1))^k=C(n,1)*2^((n-1)k)

如果至少都包含两个元素，则种数为C(n,2)*(2^(n-2))^k=C(n,2)*2^((n-2)k)

如果至少都包含i个元素，则种数为C(n,i)*(2^(n-i))^k=C(n,i)*2^((n-i)k)

减去包含一个的加上包含两个的减去包含3个的，如此类推，可以得出一下公式：

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，直接快速幂即可。

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;}