zxa and set
题目链接：
http://acm.hust.edu.cn/vjudge/contest/121332#problem/G
Description
zxa has a set , which has elements and obviously non-empty subsets.
For each subset of , which has elements, zxa defined its value as .
zxa is interested to know, assuming that represents the sum of the values of the non-empty sets, in which each set is a subset of and the number of elements in is odd, and represents the sum of the values of the non-empty sets, in which each set is a subset of and the number of elements in is even, then what is the value of , can you help him?
Input
The first line contains an positive integer , represents there are test cases.
For each test case:
The first line contains an positive integer , represents the number of the set is .
The second line contains distinct positive integers, repersent the elements .
There is a blank between each integer with no other extra space in one line.
Output
For each test case, output in one line a non-negative integer, repersent the value of .
Sample Input
3
1
10
3
1 2 3
4
1 2 3 4
Sample Output
10
3
4
Hint
题意：
给出集合A，B是A的任意子集；某个集合的权值定义为集合元素的最小值；
S-odd为元素个数为奇数的B的个数；
S-even为元素个数为偶数的B的个数；
求abs(S-odd - S-even)；
题解：
看上去好像很麻烦，要算很多次；
尝试把每个元素在结果中出现的次数写成组合数的形式；
再按题意加起来，运用组合数相关公式可以很容易得到：
结果即为A中的最大值；
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define mid(a,b) ((a+b)>>1)
#define LL long long
#define maxn 600000
#define IN freopen("in.txt","r",stdin);
using namespace std;
int n;
int main(int argc, char const *argv[])
{
//IN;
int t;cin>>t;
while(t--)
{
scanf("%d",&n);
int ans = -1;
while(n--){
int x;
cin >> x;
ans = max(ans, x);
}
printf("%d
", ans);
}
return 0;
}