There are n people who want to participate in a boat competition. The weight of the i -th participant is wi. Only teams consisting of two people can participate in this competition. As an organizer, you think that it's fair to allow only teams with the same total weight.
So, if there are k teams (a1,b1), (a2,b2), …… (ak, bk) , where ai is the weight of the first participant of the i -th team and bi is the weight of the second participant of the i -th team, then the condition a1+b1=a2+b2=...=ak+bk=s , where s is the total weight of each team, should be satisfied.
Your task is to choose such s that the number of teams people can create is the maximum possible. Note that each participant can be in no more than one team.
You have to answer t independent test cases.
Input
~~
Output
For each test case, print one integer k : the maximum number of teams people can compose with the total weight s, if you choose s optimally.
Example
Input
Copy
5
5
1 2 3 4 5
8
6 6 6 6 6 6 8 8
8
1 2 2 1 2 1 1 2
3
1 3 3
6
1 1 3 4 2 2
Output
Copy
2
3
4
1
2
这个题关键的地方就是注意到数据范围都很小（不管是n还是\(a_i\)），这就启示我们可以暴力。首先利用桶的思想预处理出来每个位置有多少个数，然后直接枚举每一个可能的重量和w，第二层循环枚举数对的其中一个数，然后直接判断桶里和这个数和为w的另一个数的个数，如果个数大于1的话说明能凑出来一对，累加到答案后更新桶里位置的个数。特别注意如果是a+a=w这种情况的话需要特判一下。
#include
using namespace std;
int a[55], ans, match[105];
bool vis[55];
int main()
{
int t;
cin >> t;
while(t--)
{
int n;
cin >> n;
ans = 0;
for(int i = 1; i <= n; i++)
{
cin >> a[i];
}
sort(a + 1, a + n +1);
for(int w = 1; w <= a[n] + a[n - 1]; w++)//枚举重量和为w
{
int cnt = 0;
memset(match, 0, sizeof(match));//重新初始化一遍桶
for(int i = 1; i <= n; i++)
{
match[a[i]]++;
}
for(int i = 1; i <= n; i++)
{
if(w - a[i] >= 1 && match[w - a[i]] && match[a[i]])//这两个数必须都得存在，因为当前枚举到的数可能已经被用过了，因此也得判断桶
{
if(w - a[i] == a[i] && match[w - a[i]] <= 1) continue;
cnt++;
match[w - a[i]]--;//更新桶
match[a[i]]--;
}
}
ans = max(ans, cnt);
}
cout << ans << endl;
}
return 0;
}