HDU 5680 zxa and set(水题)

本文介绍了一种计算特定集合所有非空子集中奇数和偶数元素数量子集价值之差的方法。通过定义子集的价值为其最小元素,利用算法快速求解|Sodd - Seven|的绝对值。

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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 (2n1) non-empty subsets.

For each subset B={b1,b2,,bm}(1mn) 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 |SoddSeven|, 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.

1T100,1n30,1ai109
 

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

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


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