HDU 5363 Key Set(快速幂)

最新推荐文章于 2021-10-19 18:21:32 发布

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本文介绍了解决HDU5363 KeySet问题的方法，该问题是关于计算集合中所有非空子集为键集的数量。键集定义为集合中整数之和为偶数的子集。文章提供了使用快速幂算法解决此问题的代码实现。

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HDU 5363 Key Set(快速幂)

Time Limit:1000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

soda has a set

with

integers

. 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

are key set.

Input

There are multiple test cases. The first line of input contains an integer

, indicating the number of test cases. For each test case:

The first line contains an integer

, 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 
      
      
       找出其子区间数，并使子区间中所有数为偶数
      
      
       子区间数=2^总区间元素个数-1
      
      
       满足条件子区间数=
       2^（总区间元素个数-1）-1
      
      
       
       #include<stdio.h>
__int64 mi(__int64 a,__int64 b)  
{  
    __int64 c=1;  
    while(b)  
    {  
        if(b&1) 
		c=c*a%1000000007;  
        b>>=1;  
        a=a*a%1000000007;  
    }  
    return c;  
}  
int main()  
{  
    int t;  
    __int64 n;  
    scanf("%d",&t);  
    while(t--)  
    {  
        scanf("%I64d",&n);  
        printf("%I64d\n",mi(2,n-1)-1);  
    }  
  
    return 0;  
}