HDU 5363 Key Set (快速幂）

最新推荐文章于 2020-11-24 12:47:06 发布

Calm微笑

最新推荐文章于 2020-11-24 12:47:06 发布

阅读量346

点赞数

CC 4.0 BY-SA版权

分类专栏：数学题快速幂

本文链接：https://blog.youkuaiyun.com/yao1373446012/article/details/52003214

数学题同时被 2 个专栏收录

141 篇文章

订阅专栏

快速幂

7 篇文章

订阅专栏

本文介绍了一个算法问题，即如何计算一个包含从1到n整数的集合中有多少个非空子集的元素之和为偶数。通过分析得出公式2^(n-1)-1，并提供了一段C语言代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Key Set

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1709 Accepted Submission(s): 905

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

Author

zimpha@zju

Source

2015 Multi-University Training Contest 6

Recommend

wange2014 | We have carefully selected several similar problems for you: 5746 5745 5744 5743 5742

题意：一个集合元素为1到n,求这个集合有多少个非空子集，并且子集的元素之和为偶数

思路：2 ^ ( n - 1) -1;

纠结为什么我用( 2 ^ n ) / 2 -1.就不对.

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