D. Many Perfect Squares

D. Many Perfect Squares

You are given a set $a_1, a_2, \ldots, a_n$ of distinct positive integers.

We define the squareness of an integer $x$ as the number of perfect squares among the numbers $a_1 + x, a_2 + x, \ldots, a_n + x$.

Find the maximum squareness among all integers $x$ between $0$ and $10^{18}$, inclusive.

Perfect squares are integers of the form $t^2$, where $t$ is a non-negative integer. The smallest perfect squares are $0, 1, 4, 9, 16, \ldots$.

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 50$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the size of the set.

The second line contains $n$ distinct integers $a_1, a_2, \ldots, a_n$ in increasing order ($1 \le a_1 < a_2 < \ldots < a_n \le 10^9$) — the set itself.

It is guaranteed that the sum of $n$ over all test cases does not exceed $50$.

Output

For each test case, print a single integer — the largest possible number of perfect squares among $a_1 + x, a_2 + x, \ldots, a_n + x$, for some $0 \le x \le 10^{18}$.

Example

input

4
5
1 2 3 4 5
5
1 6 13 22 97
1
100
5
2 5 10 17 26

output

2
5
1
2

Note

In the first test case, for $x = 0$ the set contains two perfect squares: $1$ and $4$. It is impossible to obtain more than two perfect squares.

In the second test case, for $x = 3$ the set looks like $4, 9, 16, 25, 100$, that is, all its elements are perfect squares.

 

解题思路

　　首先对于任意一个$a_i$都一定存在一个$x$使得$a_i + x$是一个完全平方数，因此答案至少是$1$。然后接下来考虑答案至少是$2$的情况，如果存在$a_i + x$和$a_j + x$均是完全平方数，那么如果最终至少存在$3$个完全平方数，就必定会包含$a_i + x$和$a_j + x$，可以发现此时$x$已经固定了，因此我们只需要枚举数组看看有多少个$a_k + x$是完全平方数就可以了。

　　$n$最大为$50$，因此可以枚举所有$a_i$和$a_j$的组合，这里假设$a_i > a_j$。那么假设有$a_i + x = n^2$，$a_j + x = m^2$，这里求解$x$并不是直接枚举$x$，而是两式做差得到$a_i - a_j = n^2 - m^2 = (n - m)(n + m)$，再把因式分解看作$(n - m)(n + m) = p \cdot q$，其中$p < q$。即有$a_i - a_j = p \cdot q$，这时就可以用试除法来求出$a_i - a_j$的因子$p$，从而得到$q = \frac{a_i - a_j}{p}$。

　　因此有 $$\begin{cases} n - m = p \\ n + m = q \end{cases}$$

　　解方程得到$n = \frac{p + q}{2}$，$m = \frac{q - p}{2}$，有解条件是$2 \mid p + q$以及$2 \mid q - p$，即$p$和$q$的奇偶性要相同。所以$x = n^2 - a_i = \left( \frac{p + q}{2} \right)^2 - a_i$，这里的$x$还要满足$0 \leq x \leq {10}^{18}$。然后再枚举一遍序列统计有多少个$a_k + x$是完全平方数就可以了。

　　这里再补充个细节。实际上$x$只需要判断是否满足$x \geq 0$就行了，不需要再判断是否满足$x \leq {10}^{18}$，因为$x$是一定不会超过${10}^{18}$，注意到$x = \left( \frac{p + q}{2} \right)^2 - a_i < \frac{(p+q)^2}{4} < (p + q)^2$，而$p + q \leq p \cdot q + 1 = a_j - a_i + 1 \leq {10}^9$，因此$x < {\left({10}^9\right)}^2 = {10}^{18}$。

　　当$p > 1$，$q > 1$，那么有$(p - 1)(q - 1) \geq 1$，即$p \cdot q - p - q + 1 \geq 1$，即$p \cdot q \geq p + q$。而当$p = 1$，那么$q = a_i - a_j$，因此$p + q = a_j - a_i + 1$。注意这里$a_i - a_j \in [1, {10}^9)$。

　　在不超过${10}^9$的数中，一个数最多有$1344$个因子，因此时间复杂度为$O(n^2 \sqrt{{10}^9} + {1344} \cdot n^3)$。

　　AC代码如下：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 60;

int a[N];

int check(LL x) {
    LL t = sqrt(x);
    return t * t == x;
}

void solve() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", a + i);
    }
    sort(a, a + n);
    int ret = 1;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            int s = a[i] - a[j];
            for (int k = 1; k <= s / k; k++) {    // 求s的因子
                if (s % k == 0) {
                    int p = k, q = s / k;
                    if (p % 2 == q % 2) {    // p和q要同奇偶性才有解
                        LL x = 1ll * (p + q) * (p + q) / 4 - a[i];
                        if (x < 0 || x > 1e18) continue;    // x要在满足的范围内
                        int t = 0;
                        for (int u = 0; u < n; u++) {    // 统计有多少个数加上x后是完全平方数
                            t += check(a[u] + x);
                        }
                        ret = max(ret, t);
                    }
                }
            }
        }
    }
    printf("%d\n", ret);
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

　　Codeforces Round #844 (Div. 1 + Div. 2, based on VK Cup 2022 - Elimination Round) A-D：https://www.cnblogs.com/BlankYang/p/17056807.html

　　Codeforces Round #844 (Div. 1 + Div. 2, based on VK Cup 2022 - Elimination Round) D (数论+思维/DP+map)：https://zhuanlan.zhihu.com/p/599387918

　　力扣第324场周赛：https://www.bilibili.com/BV1wD4y1h7Vz/