A. Phoenix and Balancetime

最新推荐文章于 2025-12-09 11:28:31 发布

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本文探讨了一个算法问题，即如何将一组按2的幂次规律排列的硬币分为两堆，使得每堆硬币的数量相等且两堆之间的重量差最小。通过示例说明了最优解的寻找过程，并提供了一段C++代码实现，但该代码存在问题，有待高手指导优化。

Phoenix has n coins with weights 2¹,2²,…,2ⁿ. He knows that n is even.He wants to split the coins into two piles such that each pile has exactly n/2 coins and the difference of weights between the two piles is minimized. Formally, let a denote the sum of weights in the first pile, and b denote the sum of weights in the second pile. Help Phoenix minimize |a−b|, the absolute value of a−b.

Input

The input consists of multiple test cases.
The first line contains an integer t (1≤t≤100) — the number of test cases.
The first line of each test case contains an integer n (2≤n≤30; n is even) — the number of coins that Phoenix has.

Output

For each test case, output one integer — the minimum possible difference of weights between the two piles.

Example

input

2
2
4

output

2
6

Note

In the first test case, Phoenix has two coins with weights 2 and 4. No matter how he divides the coins, the difference will be 4−2=2.
In the second test case, Phoenix has four coins of weight 2, 4, 8, and 16. It is optimal for Phoenix to place coins with weights 2 and 16 in one pile, and coins with weights 4 and 8 in another pile. The difference is (2+16)−(4+8)=6.
一开始没使用函数，不知道为什么会出错，求高人指点。

#include<iostream>
#include<cmath>
using namespace std;
int bal(int n)
{
 int s1,s2=0;
  for(int i=1;i<n/2;i++)
  {
   s1=s1+pow(2,i);
  }
  s1=s1+pow(2,n);
  for(int j=n/2;j<n;j++)
  {
   s2=s2+pow(2,j);
  }
  cout<<fabs(s1-s2)-n<<endl;
}
int main()
{
 int t=0;
 cin>>t;
 while(t--)
 {
  int n;
  cin>>n;
  bal(n);
 }
}