hd 5363 Key Set（快速幂）

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本文介绍了一种通过快速幂运算解决KeySet问题的方法，该问题要求计算集合S中所有元素和为偶数的非空子集的数量，并给出了一段C++代码实现。

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Key Set

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1811 Accepted Submission(s): 952

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

(1≤T≤105), indicating the number of test cases. For each test case:

The first line contains an integer

(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

Sample Output

0
1
3
7

题意 ：求真子集  即 s=2^(n-1)-1

#include<stdio.h>
__int64 quickpow(__int64 a,__int64 b)
{
	__int64 ans = 1,base = a % 1000000007;
	while(b)
	{
		if(b & 1)
		{
			ans=(base * ans) % 1000000007;
		}
		base = (base * base) % 1000000007;
		b >>= 1;
	}
	return ans;
}
int main()
{
	int t,n;
	scanf("%d",&t);
	while(t--)
	{
	    scanf("%d",&n);
	    printf("%I64d\n",quickpow(2,n-1)-1);
	}
	return 0;
}