How Many Sets I(zoj3556)

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 1
2 2

Sample Output

1
9
思路：容斥原理；
推公式；

个数为n的集合的子集有2^n个，从中选出K个使得他们的交集为空的个数。

由于集合可以重复被选，所以总的数目是2^(kn)

然后选中的集合都包含x这个数的数目是c(n,1)*2^(n-1)k

选中的集合包含x1,x2的数目是c(n,2)*2^(n-2)k

……

所以满足的集合的个数res=2^kn-c(n,1)*2^(n-1)k+c(n,2)*2(n-2)k-……

推出的公式为(2^k-1)^n
上面的讲解来自http://blog.csdn.net/qiqijianglu/article/details/7932226；

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<queue>
#include<stack>
#include<map>
#include<math.h>
using namespace std;
typedef long long LL;
LL quick(LL  n,LL m);
LL N=1e9+7;
int main(void)
{
        int i,j,k;
        LL n,m;
        while(scanf("%lld %lld",&n,&m)!=EOF)
        {
               LL cc=quick(2,m);
               cc-=1;
               cc=((cc%N)+N)%N;
               LL dd=quick(cc,n);
               printf("%lld
",dd);
        }
        return 0;
}
LL quick(LL  n,LL m)
{
    LL ans=1;
    while(m)
    {
        if(m&1)
        {
            ans=(ans*n)%N;
        }
        n=n*n%N;
        m/=2;
    }
    return ans;
}