HDU 5680 zxa and set

zxa and set

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

Total Submission(s): 189    Accepted Submission(s): 153


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

 

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给你一个集合，让求这个集合的子集中有奇数个元素的每个子集的中最小的元素之和减去有偶数个元素的每个子集的中最小的元素之和。
刚开始采用直接计算的方法，无论如何就是不对，后来才发现他们的和相互抵消，最后就是集合中的最大值。。。。。。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int t,n,i,a[11000];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
printf("%d\n",a[n-1]);
}
return 0;
}