## ZOJ3556 How Many Sets I(容斥)

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

 1 /**
2  * code generated by JHelper
4  * @author xyiyy @https://github.com/xyiyy
5  */
6
7 #include <iostream>
8 #include <fstream>
9
10 //#####################
11 //Author:fraud
12 //Blog: http://www.cnblogs.com/fraud/
13 //#####################
15 #include <iostream>
16 #include <sstream>
17 #include <ios>
18 #include <iomanip>
19 #include <functional>
20 #include <algorithm>
21 #include <vector>
22 #include <string>
23 #include <list>
24 #include <queue>
25 #include <deque>
26 #include <stack>
27 #include <set>
28 #include <map>
29 #include <cstdio>
30 #include <cstdlib>
31 #include <cmath>
32 #include <cstring>
33 #include <climits>
34 #include <cctype>
35
36 using namespace std;
37 typedef long long ll;
38
39 const ll MOD = 1000000007;
40
41 //
42 // Created by xyiyy on 2015/8/5.
43 //
44
45 #ifndef ICPC_QUICK_POWER_HPP
46 #define ICPC_QUICK_POWER_HPP
47 typedef long long ll;
48
49 ll quick_power(ll n, ll m, ll mod) {
50     ll ret = 1;
51     while (m) {
52         if (m & 1) ret = ret * n % mod;
53         n = n * n % mod;
54         m >>= 1;
55     }
56     return ret;
57 }
58
59 #endif //ICPC_QUICK_POWER_HPP
60
62 public:
63     void solve(std::istream &in, std::ostream &out) {
64         int n, k;
65         while (in >> n >> k) {
66             ll ans = quick_power(2, k, MOD) - 1;
67             ans = (ans + MOD) % MOD;
68             ans = quick_power(ans, n, MOD);
69             out << ans << endl;
70         }
71     }
72 };
73
74 int main() {
75     std::ios::sync_with_stdio(false);
76     std::cin.tie(0);
78     std::istream &in(std::cin);
79     std::ostream &out(std::cout);
80     solver.solve(in, out);
81     return 0;
82 }

posted on 2015-08-16 01:05  xyiyy  阅读(...)  评论(...编辑  收藏

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