HDU 5680 zxa and set（水题）

思路：其实就是求最大值...

#include<bits/stdc++.h>
using namespace std;

void solve()
{
    int n;scanf("%d",&n);
    int ans=0,x;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        ans=max(ans,x);
    }
    printf("%d\n",ans);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)solve();
    return 0;
}

Problem Description

zxa has a set A={a1,a2,⋯,an}, which has n elements and obviously (2n−1) non-empty subsets.

For each subset B={b1,b2,⋯,bm}(1≤m≤n) of A, which has m elements, zxa defined its value as min(b1,b2,⋯,bm).

zxa is interested to know, assuming that Sodd represents the sum of the values of the non-empty sets, in which each set B is a subset of A and the number of elements in B is odd, and Seven represents the sum of the values of the non-empty sets, in which each set B is a subset of A and the number of elements in B is even, then what is the value of |Sodd−Seven|, can you help him?

 

Input

The first line contains an positive integer T, represents there are T test cases.

For each test case:

The first line contains an positive integer n, represents the number of the set A is n.

The second line contains n distinct positive integers, repersent the elements a1,a2,⋯,an.

There is a blank between each integer with no other extra space in one line.

1≤T≤100,1≤n≤30,1≤ai≤109

 

Output

For each test case, output in one line a non-negative integer, repersent the value of |Sodd−Seven|.

 

Sample Input

3
1
10
3
1 2 3
4
1 2 3 4

 

Sample Output

10
3
4

Hint
For the first sample, $A=\{10\}$, which contains one subset $\{10\}$ in which the number of elements is odd, and no subset in which the number of elements is even, therefore $S_{odd}=10,S_{even}=0,|S_{odd}-S_{even}|=10$.

For the second sample, $A=\{1,2,3\}$, which contains four subsets $\{1\},\{2\},\{3\},\{1,2,3\}$ in which the number of elements is odd, and three subsets $\{1,2\},\{2,3\},\{1,3\}$ in which the number of elements is even, therefore $S_{odd}=1+2+3+1=7,S_{even}=1+2+1=4,|S_{odd}-S_{even}|=3$.