preview round guarantee () == ecif sam different n-1

B. Mike and Children time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Mike decided to teach programming to children in an elementary school. He knows that it is not an easy task to interest children in that age to code. That is why he decided to give each child two sweets.

Mike has nn sweets with sizes a1,a2,…,ana1,a2,…,an . All his sweets have different sizes. That is, there is no such pair (i,j)(i,j) (1≤i,j≤n1≤i,j≤n ) such that i≠ji≠j and ai=ajai=aj .

Since Mike has taught for many years, he knows that if he gives two sweets with sizes

aiai and ajaj to one child and akak and apap to another, where (ai+aj)≠(ak+ap)(ai+aj)≠(ak+ap) , then a child who has a smaller sum of sizes will be upset. That is, if there are two children who have different sums of sweets, then one of them will be upset. Apparently, Mike does not want somebody to be upset.

Mike wants to invite children giving each of them two sweets. Obviously, he can‘t give one sweet to two or more children. His goal is to invite as many children as he can.

Since Mike is busy preparing to his first lecture in the elementary school, he is asking you to find the maximum number of children he can invite giving each of them two sweets in such way that nobody will be upset.

Input

The first line contains one integer nn (2≤n≤10002≤n≤1000 ) — the number of sweets Mike has.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1051≤ai≤105 ) — the sizes of the sweets. It is guaranteed that all integers are distinct.

Output

Print one integer — the maximum number of children Mike can invite giving each of them two sweets in such way that nobody will be upset.

Examples Input Copy

8
1 8 3 11 4 9 2 7

Output Copy

3

Input Copy

7
3 1 7 11 9 2 12

Output Copy

2

Note

In the first example, Mike can give 9+2=119+2=11 to one child, 8+3=118+3=11 to another one, and 7+4=117+4=11 to the third child. Therefore, Mike can invite three children. Note that it is not the only solution.

In the second example, Mike can give 3+9=123+9=12 to one child and 1+111+11 to another one. Therefore, Mike can invite two children. Note that it is not the only solution.

這個題目，其實不太難，但是我自己還是沒有獨立解決，水平嚴重有待提升，這個呢要註意n個數都互不相等的，所以這說明要是求出相等的，則肯定不會用到同一個數 （a[i]+a[j]==a[i]+a[k]不可能成立）

知道這個我就會寫了，就直接二重循環即可

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
const int maxn = 1300;
const int inf=2e5+10;
int p[inf];
bool cmp(int a,int b)
{
	return a>b;
}
int main()
{
	int n,a[maxn];
	ll ans=0;
	memset(p, 0, sizeof(p));
	cin >> n;
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	for (int i = 1; i <= n; i++)
	{
		for (int j = i+1; j <= n; j++)
		{
			p[a[i] + a[j]]++;
			ans=max(1ll*p[a[i]+a[j]],ans);
		}
		
	}
	printf("%d\n",ans);
	return 0;
}

　　

cf 20190307 Codeforces Round #543 (Div. 2, based on Technocup 2019 Final Round)