BestCoder Round #83 1001&&HDU-5680 zxa and set (水)

BC #83 1001（HDU-5680） zxa and set (水)

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}A=\{10\}, which contains one subset {10}\{10\} in which the number of elements is odd, and no subset in which the number of elements is even, therefore Sodd=10,Seven=0,|Sodd−Seven|=10S_{odd}=10,S_{even}=0,|S_{odd}-S_{even}|=10.

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

题意：找个数为奇数与个数为偶数的集合（子集中的 所有最小值） 之和 之差，其实根据推算俩式子一划，就是寻找原集合中最大值。

#include"iostream"
#include"cstdio"
using namespace std;
int main(){
int t,n,m,j,i;
long long a[20010],sum,x,y;
scanf("%d",&t);
while(t--){
scanf("%d %d",&n,&m);
for(i=1; i<=n; i++){
scanf("%lld",&a[i]);
}
sum = 0;
for(i=1; i<n; i++){
for(j=i+1; j<=n; j++)
sum ^= (a[i]+a[j]);
}
while(m--){
scanf("%lld %lld",&x,&y);
for(i=1; i<=n; i++){
if(i!=x){
sum ^=(a[i]+y),sum^=(a[i]+a[x]);
}

}
printf("%lld\n",sum);
a[x] = y;
}
}
return 0;
}