Even Subarrays（数论问题）

题目链接

题目描述：

You are given an integer array \(a_1,a_2,…,a_n (1≤a_i≤n).\)

Find the number of subarrays of a whose XOR has an even number of divisors. In other words, find all pairs of indices \((i,j) (i≤j)\) such that \(a_i⊕a_{i+1}⊕⋯⊕a_j\) has an even number of divisors.

For example, numbers \(2, 3, 5\) or \(6\) have an even number of divisors, while \(1\) and \(4\) — odd. Consider that 0 has an odd number of divisors in this task.

Here XOR (or \(⊕\)) denotes the bitwise XOR operation.

Print the number of subarrays but multiplied by 2022... Okay, let's stop. Just print the actual answer.

输入描述：

Each test contains multiple test cases. The first line contains the number of test cases \(t (1≤t≤10^4)\). Description of the test cases follows.

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

The second line contains n integers \(a_1,a_2,…,a_n (1≤a_i≤n)\).

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

输入描述：

For each test case, print the number of subarrays, whose XOR has an even number of divisors.

样例：

input：

4
3
3 1 2
5
4 2 1 5 3
4
4 4 4 4
7
5 7 3 7 1 7 3

output：

4
11
0
20

AC代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

int T = 1;
LL n;

void solve()
{
    scanf("%lld", &n);

    vector<int> a(n), b(2 * n); // 开两个vector，一个用于存数组，一个用于存某个数出现的次数

    a.clear(), b.clear();

    for (int i = 0; i < n; i++)
    {
        scanf("%d", &a[i]);
    }

    LL cnt = 0; // 存储异或和的因数为奇数个的数的个数
    int t = 0;

    b[t]++; // a[0] ^ a[0]为0

    for (int i = 0; i < n; i++)
    {
        t ^= a[i];
        // cout << t << ' ';
        // n个数可能的异或最大值肯定小于2 * n
        // 判断某个数的因数是否为奇数的条件就是判断其是否是完全平方数
        for (LL j = 0; j * j < 2 * n; j++)
        {
            // 一个数 t 异或一个完全平方数的结果，该结果异或 t的结果必定是那个完全平方数
            if ((t ^ (j * j)) < 2 * n)
            {
                cnt += b[t ^ (j * j)]; // cnt 加上这个数出现过的次数

                cout << t << ' ' << j * j << ' ' << (t ^ (j * j)) << ' ' << cnt << '\n';
            }
        }

        b[t]++; // t的出现的次数加一
    }

    LL ans = (n * (n + 1) / 2) - cnt; // 总共的个数减去不成立的个数即为成立的个数

    printf("%lld\n", ans);
}

int main()
{
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

    scanf("%d", &T);

    while (T--)
    {
        solve();
    }

    return 0;
}