HDU 5363 Key Set(2015 Multi-University Training Contest 6 2015多校联合)

Key Set

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 476    Accepted Submission(s): 251

Problem Description

soda has a set S with n integers {1,2,…,n}. A set is called key set if the sum of integers in the set is an even number. He wants to know how many nonempty subsets of S are key set.

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤105), indicating the number of test cases. For each test case:The first line contains an integer n (1≤n≤109), the number of integers in the set.

 

Output

For each test case, output the number of key sets modulo 1000000007.

 

Sample Input

41234

 

Sample Output

0137

 

题目传送门：HDU 5363 Key Set

规律很容易找到。ans=pow(2,n-1)-1。但是pow到2^62就会爆。所以方法就是改写一下pow函数。然后在计算过程中%MOD。

用高精度幂运算模板过就行了。

注意输入1的时候，代码会爆栈。所以要用一个if。

代码：

#include <cstdio>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;long long mypow(long long x, long long y){if (y == 1) return x;long long int result = 0;long long int tmp = mypow(x, y / 2) % 1000000007;if (y & 1 != 0){result = (x * tmp * tmp) % 1000000007;}else{result = (tmp * tmp) % 1000000007;}return result;}int main(){long long T,n;scanf("%lld", &T);while(T--){scanf("%lld", &n);if (n == 1){printf("0\n");continue;}printf("%lld\n",mypow(2,n-1)% 1000000007-1);}}