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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
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
1 /** 2 * code generated by JHelper 3 * More info: https://github.com/AlexeyDmitriev/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 //##################### 14 //#pragma comment(linker, "/STACK:102400000,102400000") 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 61 class TaskH { 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); 77 TaskH solver; 78 std::istream &in(std::cin); 79 std::ostream &out(std::cout); 80 solver.solve(in, out); 81 return 0; 82 }