How Many Sets I(zoj3556)

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, ..., 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；

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<queue>
 6 #include<stack>
 7 #include<map>
 8 #include<math.h>
 9 using namespace std;
10 typedef long long LL;
11 LL quick(LL  n,LL m);
12 LL N=1e9+7;
13 int main(void)
14 {
15         int i,j,k;
16         LL n,m;
17         while(scanf("%lld %lld",&n,&m)!=EOF)
18         {
19                LL cc=quick(2,m);
20                cc-=1;
21                cc=((cc%N)+N)%N;
22                LL dd=quick(cc,n);
23                printf("%lld\n",dd);
24         }
25         return 0;
26 }
27 LL quick(LL  n,LL m)
28 {
29     LL ans=1;
30     while(m)
31     {
32         if(m&1)
33         {
34             ans=(ans*n)%N;
35         }
36         n=n*n%N;
37         m/=2;
38     }
39     return ans;
40 }