Key Set<hdoj5363>

最新推荐文章于 2022-07-20 11:50:38 发布

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最新推荐文章于 2022-07-20 11:50:38 发布

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本文介绍了一种算法，用于计算给定整数集合中所有非空子集中，元素之和为偶数的子集数量，并提供了相应的代码实现。输入包含多个测试用例，每个用例给出一个整数集合的大小，输出则是满足条件的子集数量模1000000007的结果。

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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

#include<cstdio>
long long quickpow(long long n)
{
long long ans=1,base=2;
     n=n-1;
	while(n)
	{
		if(n&1)
		{
			ans=(base*ans)%1000000007;
		}
		base=(base*base)%1000000007;
		n>>=1;
	}
	return ans-1;
}
int main()
{
     int t;
     scanf("%d",&t);
	 while(t--)
	 {
	 	long long n;
	 	scanf("%lld",&n);
	 	printf("%lld\n",quickpow(n));
	  } 
	return 0;
}