codeforces 1348A - Phoenix and Balance

A. Phoenix and Balance time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output 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 n2 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 inputCopy 2 2 4 outputCopy 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.
リンク:https://codeforces.ml/problemset/problem/1348/A題意:21-2 nは2つの山に分けて、2つの山の相殺するminの構想を求めます:等比数列an>sn-1(q>1)(自推、コメントが必要なら私は更に導出の過程をあげます)によって最後の1項がきっと前のすべての数より加算するのが大きいことを知ることができますので、2つの山の相殺するminを求めて直接前のn/2-1項+最後の1項-残りの数acコードは以下の通りです:

#include
using namespace std;
int main()
{
	int a[31]={2,4};
	for(int i=2;i<31;i++)
	{
		a[i] = 2*a[i-1];
	}
	int t;
	cin>>t;
	while(t--)
	{
		int n;
		cin>>n;
		int sum1=0,sum2=0; 
		
		for(int i=0;i