ZOJ3556 How Many Sets I(容斥)

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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

容斥推一下公式，大致就是有一个相交，两个相交。。。。。。

最后可以推出公式是((2^k)-1)^n，还是比较容易得出公式的

 /**
  * code generated by JHelper
  * More info: https://github.com/AlexeyDmitriev/JHelper
  * @author xyiyy @https://github.com/xyiyy
  */

 #include <iostream>
 #include <fstream>

 //#####################
 //Author:fraud
 //Blog: http://www.cnblogs.com/fraud/
 //#####################
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include <iostream>
 #include <sstream>
 #include <ios>
 #include <iomanip>
 #include <functional>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <list>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <set>
 #include <map>
 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <climits>
 #include <cctype>

 using namespace std;
 typedef long long ll;

 const ll MOD = ;

 //
 // Created by xyiyy on 2015/8/5.
 //

 #ifndef ICPC_QUICK_POWER_HPP
 #define ICPC_QUICK_POWER_HPP
 typedef long long ll;

 ll quick_power(ll n, ll m, ll mod) {
     ll ret = ;
     while (m) {
         if (m & ) ret = ret * n % mod;
         n = n * n % mod;
         m >>= ;
     }
     return ret;
 }

 #endif //ICPC_QUICK_POWER_HPP

 class TaskH {
 public:
     void solve(std::istream &in, std::ostream &out) {
         int n, k;
         while (in >> n >> k) {
             ll ans = quick_power(, k, MOD) - ;
             ans = (ans + MOD) % MOD;
             ans = quick_power(ans, n, MOD);
             out << ans << endl;
         }
     }
 };

 int main() {
     std::ios::sync_with_stdio(false);
     std::cin.tie();
     TaskH solver;
     std::istream &in(std::cin);
     std::ostream &out(std::cout);
     solver.solve(in, out);
     return ;
 }