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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
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
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;
}
hdu 5680 zxa and set(水题)
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原文地址：http://blog.csdn.net/qiqi_skystar/article/details/51812319
