ZOJ 3556 How Many Sets I

How Many Sets I

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on ZJU. Original ID: 3556
64-bit integer IO format: %lld      Java class name: Main

 

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

Source

ZOJ Monthly, October 2011

Author

QU, Zhe

 

解题：

从子集中选k个的有序组合个数（子集可重复被选中）有 $(2^n)^k=2^{n imes k}$;

 故总数为$S=2^{n imes k}$;

 $设S(x)为k个集合的有序组合的个数，这些集合都包含至少一个x。$

 $S(x_1 & x_2)为k个集合的有序组合的个数，这些集合都包含至少一个x_1和x_2$

 $S(x_1& x_2& x_3...x_k)为k个集合的有序组合的个数，这些集合都包含至少一个x_1,x_2...x_k。$

 $而  S(x)=(2^{n-1})^k =2^{(n-1) imes k};$

 $S(x_1 & x_2)=(2^{n-2})^k=2^{(n-2) imes k};$

 $S(x_1& x_2 &...& x_i)=(2^{n-i})^k=2^{(n-i) imes k};$

 由容斥原理知，我们要得到的就是

$S-(n,1) imes S(x) + (n,2) imes (S(x_1& x_2)-(n,3) imes (S(x_1& x_2&x_3)+....(-1)^i imes (n,i) imes S(x_1& x_2&...x_i)...(-1)^n imes (n,n) imes S(x_1& x_2...& x_n);$

$化简得ans=(2^K-1)^N;$

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL mod = 1000000007,n,k;
LL quickPow(LL base,LL index){
    LL ret = 1;
    while(index){
        if(index&1) ret = ret*base%mod;
        index >>= 1;
        base = base*base%mod;
    }
    return ret;
}
int main(){
    while(~scanf("%lld%lld",&n,&k))
        printf("%lld
",quickPow(((quickPow(2,k) - 1)%mod + mod)%mod,n));
    return 0;
}