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 = 1000000007;

//
// 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 = 1;
    while (m) {
        if (m & 1) ret = ret * n % mod;
        n = n * n % mod;
        m >>= 1;
    }
    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(2, k, MOD) - 1;
            ans = (ans + MOD) % MOD;
            ans = quick_power(ans, n, MOD);
            out << ans << endl;
        }
    }
};

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    TaskH solver;
    std::istream &in(std::cin);
    std::ostream &out(std::cout);
    solver.solve(in, out);
    return 0;
}