Equalize

题面翻译

有一个给定的长度为 $n$ 的数列 $a$，现在加上一个排列 $b$，即 $c_i=a_i+b_i$。

现在求对于所有可能的 $b$，$c$ 中出现最多的数的出现次数的最大值。

translate by @UniGravity.

题目描述

Vasya has two hobbies — adding permutations $^{\dagger}$ to arrays and finding the most frequently occurring element. Recently, he found an array $a$ and decided to find out the maximum number of elements equal to the same number in the array $a$ that he can obtain after adding some permutation to the array $a$ .

More formally, Vasya must choose exactly one permutation $p_1, p_2, p_3, \ldots, p_n$ of length $n$ , and then change the elements of the array $a$ according to the rule $a_i := a_i + p_i$ . After that, Vasya counts how many times each number occurs in the array $a$ and takes the maximum of these values. You need to determine the maximum value he can obtain.

$^{\dagger}$ A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ( $2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ( $n=3$ but there is $4$ in the array).

输入格式

Each test consists of multiple test cases. The first line contains a single integer $t$ ( $1 \leq t \leq 2 \cdot 10^4$ ) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ( $1 \le n \le 2 \cdot 10^5$ ) — the length of the array $a$ .

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $1 \le a_i \le 10^9$ ) — the elements of the array $a$ .

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$ .

输出格式

For each test case, output a single number — the maximum number of elements equal to the same number after the operation of adding a permutation.

样例 #1

样例输入 #1

7
2
1 2
4
7 1 4 1
3
103 102 104
5
1 101 1 100 1
5
1 10 100 1000 1
2
3 1
3
1000000000 999999997 999999999

样例输出 #1

2
2
3
2
1
1
2

提示

In the first test case, it is optimal to choose $p = [2, 1]$ . Then after applying the operation, the array $a$ will be $[3, 3]$ , in which the number $3$ occurs twice, so the answer is $2$ .

In the second test case, one of the optimal options is $p = [2, 3, 1, 4]$ . After applying the operation, the array $a$ will be $[9, 4, 5, 5]$ . Since the number $5$ occurs twice, the answer is $2$ .

如果一个区间的右端点和左端点的差值小于n，那么这个区间就可以用排列的数字凑成数值一样的区间，找最长的区间

首先从小到大排列，然后去重，去重的意义是排列的数字都是不同的，对于同一个数字加上不同的数一定不等，所一相同的数字取一个即可

const int N = 2e5 + 10;
int a[N];
int b[N];
void solve()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    sort(a + 1, a + 1 + n);
    int h = 0;
    for (int i = 1; i <= n; i++)
    {
        if (a[i] != a[i - 1]) b[++h] = a[i];
    }
    int i = 1;
    int j = i + 1;
    int ans = 0;
    while (i <= h)
    {
        while (b[j] - b[i] < n && j <= h) j++;
        ans = max(ans, j - i);
        i++;
    }
    cout << ans << '\n';
}