题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5680

zxa and set

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 411    Accepted Submission(s): 310

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 $S_{odd}=1+2+3+1=7,S_{even}=1+2+1=4,|S_{odd}-S_{even}|=3$.

Source

BestCoder Round #83

题目大意：简答的说就是求最大值~~~~~

详见代码。

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

int a[1010];

int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int n;
        scanf("%d",&n);
        for (int i=0;i<n;i++)
        {
            scanf ("%d",&a[i]);
        }
        sort(a,a+n);
        printf("%d\n",a[n-1]);
    }
    return 0;
}