Codeforces Round 948 (Div. 2) C. Nikita and LCM 题解 数学

Nikita and LCM

题目描述

Nikita is a student passionate about number theory and algorithms. He faces an interesting problem related to an array of numbers.

Suppose Nikita has an array of integers $a$ of length $n$. He will call a subsequence${}^{†}$ of the array special if its least common multiple (LCM) is not contained in $a$. The LCM of an empty subsequence is equal to $0$.

Nikita wonders: what is the length of the longest special subsequence of $a$? Help him answer this question!

${}^{†}$ A sequence $b$ is a subsequence of $a$ if $b$ can be obtained from $a$ by the deletion of several (possibly, zero or all) elements, without changing the order of the remaining elements. For example, $[5,2,3]$ is a subsequence of $[1,5,7,8,2,4,3]$.

输入描述

Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1\le t\le 2000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 2000$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

输出描述

For each test case, output a single integer — the length of the longest special subsequence of $a$.

样例输入 #1

6
5
1 2 4 8 16
6
3 2 10 20 60 1
7
2 3 4 6 12 100003 1200036
9
2 42 7 3 6 7 7 1 6
8
4 99 57 179 10203 2 11 40812
1
1

样例输出 #1

0
4
4
5
8
0

原题

CF——传送门
洛谷——传送门

思路&代码

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

int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
int lcm(int a, int b)
{
    return a / gcd(a, b) * b;
}

void solve()
{
    int n;
    cin >> n;
    vector<int> a(n);
    int ans = 0;
    map<int, int> mp; // map记录存在在数组a中的数
    int maxNum = 0;   // 我们需要先找到最大值
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        mp[a[i]] = 1;
        maxNum = max(maxNum, a[i]);
    }
    for (int i = 0; i < n; i++)
    {
        // 如果数组中有一个数不整除最大值，那么二者的lcm一定大于最大值，即一定不在数组a中，那么最长的满足条件的子序列长度为n
        if (maxNum % a[i] != 0)
        {
            cout << n << '\n';
            return;
        }
    }
    // 我们需要知道，一旦该数组不满足上面的这种情况(即ans为n的情况)，那么该数组就有存在一个性质
    // 即：数组中的任何一个数都是最大值的因数(条件1)
    // 于是我们可以知道，数组中任意取数进行lcm运算，所得到的值一定也是最大值的因数
    // 那么我们的策略就是，找到最大值的所有因数，如果该数不在数组a中，则检查其是否能由数组中的数进行lcm运算得到，如果能得到，更新ans为所有可能答案中的最大值
    auto updata = [&](int subsequence_lcm) // 更新满足条件的子序列中的最长长度
    {
        int cur_lcm = 1; // 当前在数组中的所有subsequence_lcm的因数通过lcm运算得到的值
        int cnt = 0;     // subsequence_lcm的因数个数
        for (int i = 0; i < n; i++)
        {
            if (subsequence_lcm % a[i] == 0)
            {
                cnt++;
                cur_lcm = lcm(cur_lcm, a[i]);
            }
        }
        if (cur_lcm == subsequence_lcm) // 如果subsequence_lcm能由数组中的数通过lcm运算得到，则其是满足条件的一种情况
            ans = max(ans, cnt);
    };
    for (int i = 1; i * i <= maxNum; i++) // 利用数学知识，可由maxNum/i运算得到maxNum的另一个因数，从而降低遍历复杂度
    {
        if (maxNum % i == 0)
        {
            if (mp[i] == 0)
                updata(i);
            if (mp[maxNum / i] == 0)
                updata(maxNum / i);
        }
    }
    cout << ans << '\n';
}

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

    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}