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

4 1 2 3 4

 

Sample Output

0 1 3 7

 

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 
#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);
	}
}