CF1121B.Mike and Children

普及/提高-

通过率:0%

AC君温馨提醒

该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。

题目描述

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,,ana_1, a_2, \ldots, a_n . All his sweets have different sizes. That is, there is no such pair (i,j)(i, j) ( 1i,jn1 \leq i, j \leq n ) such that iji \ne j and ai=aja_i = a_j .

Since Mike has taught for many years, he knows that if he gives two sweets with sizes aia_i and aja_j to one child and aka_k and apa_p to another, where (ai+aj)(ak+ap)(a_i + a_j) \neq (a_k + a_p) , 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.

输入格式

The first line contains one integer nn ( 2n10002 \leq n \leq 1\,000 ) — the number of sweets Mike has.

The second line contains nn integers a1,a2,,ana_1, a_2, \ldots, a_n ( 1ai1051 \leq a_i \leq 10^5 ) — the sizes of the sweets. It is guaranteed that all integers are distinct.

输出格式

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.

输入输出样例

  • 输入#1

    8
    1 8 3 11 4 9 2 7
    

    输出#1

    3
    
  • 输入#2

    7
    3 1 7 11 9 2 12
    

    输出#2

    2
    

说明/提示

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.

首页