钉耙春季2025第十场——小塔的序列

求最长乘积为完全平方数的连续子序列。一个数位完全平方数等价于分解质因数后每一个质因数的数量位偶数。先将所有1e6以内的质数哈希，防止出现1²3==0的情况出现，使异或和为零必须是每个数出现的次数为偶数。
将ai分解质因数，求一个ai->bi的映射，bi定义为ai的哈希后的质因数的异或和。对bi求前缀和，前缀和中相离最远的两个相同的数的距离就是答案

```cpp
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <vector>
#include <map>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <functional>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <iomanip>
#include <random>
#include <chrono>

#define int long long
#define INF 1e18
#define endl '\n'
#define MOD (int)(1000000007)

using namespace std;

int n, m;
vector<int> arr(1000006+100);
vector<map<int, int>> num_prim(1000006 + 100);
vector<int> primes;
map<int, size_t> int2hash;

vector<size_t> sumhash(1000006 + 100);


static mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());

unordered_map<size_t, size_t>hashmap;
size_t hash_int(int x) 
{
    if(hashmap[x]==0)
    {
        hashmap[x]=rnd();
        return hashmap[x];
    }
    else
    {
        return hashmap[x];
    }
}
// hash<int> hash_int;


void solve()
{
    int n;
    cin >> n;
    vector<int> a(n + 5, 0);
    vector<size_t> b(n + 5);

    map<size_t, int> num_bd;
    size_t nowhash = 0;

    long long ans = 0;
    long long ansl = -1, ansr = -1;
    num_bd[nowhash] = 0;

    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        b[i] = sumhash[a[i]];
        if (b[i] == 0 && ans == 0)
        {
            ans = 1;
            ansl = i;
            ansr = i;
        }
        nowhash = nowhash ^ b[i];
        int nowindex = num_bd[nowhash];
        if (nowhash == 0)
        {
            if (i > ans)
            {
                ansl = 1;
                ansr = i;
                ans = i;
            }
        }
        else if (nowindex == 0)
        {
            num_bd[nowhash] = i;
        }
        else
        {
            if (i - nowindex > ans)
            {
                ansl = nowindex + 1;
                ansr = i;
                ans = i - nowindex;
            }
            else if (i - nowindex == ans && nowindex + 1 < ansl)
            {
                ansl = nowindex + 1;
                ansr = i;
            }
        }
    }
    if (ans <= 0)
    {
        cout << "-1 -1" << endl;
    }
    else
    {
        cout << ansl << " " << ansr << endl;
    }
}

signed main()
{
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);

    for (int i = 2; i <= 1e6 + 50; i++)
    {
        if (arr[i] == 0)
        {
            int nowhash = hash_int(i);
            for (int j = i; j <= 1e6 + 50; j += i)
            {
                arr[j] = 1;
                int cpj = j;
                while (cpj % i == 0 && cpj != 0)
                {
                    sumhash[j] ^= nowhash;
                    cpj /= i;
                }
            }
        }
    }
    sumhash[1] = 0;

    int T = 1;
    cin >> T;
    for (int i = 1; i <= T; i++)
    {
        solve();
    }
    return 0;
}

收获：
1.O(nlogn)分解质因数：埃氏筛遍历小于等于值域的当前质因数的倍数，while(j%i==0){j/=i,cnt++};求出以j由多少个i组成。
2.用哈希异或和求出每个类别数量和是否为偶数