2025.1.19——1300

合集 - 日常训练(56)

1.2025.2.19——150002-192.11.23 周六2024-11-243.11.24 周日2024-11-254.11.25 周一日常2024-11-265.2024.11.26 周二日常2024-11-266.2024.11.27 周三2024-11-277.2024.11.28周四2024-11-288.2024.11.29 周五2024-11-299.2024.11.30 周六2024-11-3010.2024.12.1 周日2024-12-0211.2024.12.2 周一2024-12-0212.2024.12.3 周二2024-12-0413.2024.12.4 周三2024-12-0514.2024.12.5 周四2024-12-0615.2024.12.7 周六2024-12-0816.2024.12.8 周日2024-12-0917.2024.12.9 周一2024-12-1018.2024.12.10 周二2024-12-1119.2024.12.11 周三2024-12-1220.2024.12.12 周四2024-12-1321.2024.12.13 周五2024-12-1422.2024.12.14 周六2024-12-1523.2024.12.16 周一2024-12-1724.2024.12.17 周二2024-12-1825.2024.12.18 周三2024-12-1926.2024.12.19 周四2024-12-2027.2024.12.20 周五2024-12-2128.2024.12.21 周六2024-12-2229.2024.12.22 周日2024-12-2330.2024.12.23 周一2024-12-2431.2024.12.24 周四2024-12-2532.2024.12.25 周三2024-12-2633.2024.12.26 周四2024-12-2834.2024.12.27 周五2024-12-2835.2024.12.28 周六2024-12-2936.2024.12.29 周日2024-12-3037.2024.12.30 周一01-0138.2025.1.5——120001-1439.2025.1.12——120001-1440.2025.1.14——120001-1541.2025.1.15——120001-1642.2025.1.16——120001-1743.2025.1.17——120001-1844.2025.1.18——130001-19

45.2025.1.19——130001-20

46.2025.1.20——130001-2147.2025.1.21——130001-2248.2025.1.22——130001-2449.2025.1.24——140001-2650.2025.1.26——140001-2851.2025.2.8——140002-0952.2025.2.9——140002-1153.2025.2.10——140002-1154.2025.2.14——140002-1455.2025.2.15——140002-1556.2025.2.17——140002-17

收起

2025.1.19——1300

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A 1300

You are given an array $a$ consisting of $n$ positive integers. You can perform the following operation on it:

1. Choose a pair of elements $a_i$ and $a_j$ ($1\le i,j\le n$ and $i\ne j$);
2. Choose one of the divisors of the integer $a_i$, i.e., an integer $x$ such that $a_{i}modx=0$;
3. Replace $a_i$ with $\frac{a_i}{x}$ and $a_j$ with $a_j\cdot x$.

Determine whether it is possible to make all elements in the array the same by applying the operation a certain number of times (possibly zero).

For example, let's consider the array $a$ = [$100,2,50,10,1$] with $5$ elements. Perform two operations on it:

1. Choose $a_3=50$ and $a_2=2$, $x=5$. Replace $a_3$ with $\frac{a_3}{x}=\frac{50}{5}=10$, and $a_2$ with $a_2\cdot x=2\cdot 5=10$. The resulting array is $a$ = [$100,10,10,10,1$];
2. Choose $a_1=100$ and $a_5=1$, $x=10$. Replace $a_1$ with $\frac{a_1}{x}=\frac{100}{10}=10$, and $a_5$ with $a_5\cdot x=1\cdot 10=10$. The resulting array is $a$ = [$10,10,10,10,10$].

After performing these operations, all elements in the array $a$ become equal to $10$.
Input

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

Then follows the description of each test case.

The first line of each test case contains a single integer $n$ ($1\le n\le {10}^4$) — the number of elements in the array $a$.

The second line of each test case contains exactly $n$ integers $a_i$ ($1\le a_i\le {10}^6$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed ${10}^4$.

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B 1300

You have $n$ sets of integers $S_1,S_2,\dots ,S_n$. We call a set $S$ attainable, if it is possible to choose some (possibly, none) of the sets $S_1,S_2,\dots ,S_n$ so that $S$ is equal to their union${}^{†}$. If you choose none of $S_1,S_2,\dots ,S_n$, their union is an empty set.

Find the maximum number of elements in an attainable $S$ such that $S\ne S_1\cup S_2\cup \dots \cup S_n$.

${}^{†}$ The union of sets $A_1,A_2,\dots ,A_k$ is defined as the set of elements present in at least one of these sets. It is denoted by $A_1\cup A_2\cup \dots \cup A_k$. For example, $\{2,4,6\}\cup \{2,3\}\cup \{3,6,7\}=\{2,3,4,6,7\}$.
Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). 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 following $n$ lines describe the sets $S_1,S_2,\dots ,S_n$. The $i$-th of these lines contains an integer $k_i$ ($1\le k_i\le 50$) — the number of elements in $S_i$, followed by $k_i$ integers $s_{i,1},s_{i,2},\dots ,s_{i,k_i}$ ($1\le s_{i,1}<s_{i,2}<\dots <s_{i,k_i}\le 50$) — the elements of $S_i$.

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------------------------思考------------------------

• 数学+思维/入手点

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A

1. 本质操作是移动因子/质因子。
2. 从结果来看，分解质因数后发现每个质因子出现的次数都必须为 $n$ 的倍数。具有充分必要性。

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B

1. 入手点是枚举每个数：当此数不在最后集合中时，暴力统计有多少元素没有受到影响。最后坏情况大约 6e7 , 600ms 。
2. 搞了半天数据结构维护，原来暴力就行。

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------------------------代码------------------------

A

#include <bits/stdc++.h>
#define int long long //
#define endl '\n'     // attention: interactive/debug
#define el cout << endl
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
#define bugv(VEC, ST)                     \
    for (int i = ST; i < VEC.size(); i++) \
        cout << VEC[i] << ' ';            \
    el;

void _();
signed main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cout << fixed << setprecision(10);
    int T = 1;
    cin >> T;
    while (T--)
        _();
    return 0;
}

void _()
{
    int n;
    cin >> n;
    map<int, int> cnt;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        for (int j = 2; j <= x / j && x; j++)
            while (x % j == 0)
            {
                cnt[j]++;
                x /= j;
            }
        if (x - 1)
            cnt[x]++;
    }
    // for (auto [x, has] : cnt)
    //     bug2(x, has);
    bool f = 1;
    for (auto [x, has] : cnt)
        if (has % n)
            f = 0;
    cout << (f ? "YES" : "NO");
    el;
}

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B

#include <bits/stdc++.h>
#define int long long //
#define endl '\n'     // attention: interactive/debug
#define el cout << endl
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
#define bugv(VEC, ST)                                    \
    for (auto i = VEC.begin() + ST; i != VEC.end(); i++) \
        cout << *i << ' ';                               \
    el;

void _();
signed main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cout << fixed << setprecision(10);
    int T = 1;
    cin >> T;
    while (T--)
        _();
    return 0;
}

void _()
{
    int n;
    cin >> n;
    vector<set<int>> has(n);
    set<int> t;

    for (auto &s : has)
    {
        int cnt;
        cin >> cnt;
        for (int i = 0; i < cnt; i++)
        {
            int x;
            cin >> x;
            s.insert(x);
            t.insert(x);
        }
    }

    n = t.size();
    int res = 0;
    for (auto x : t)
    {
        set<int> left;
        for (auto s : has)
        {
            auto it = s.lower_bound(x);
            // bug(*it);
            if (it == s.end() || *it != x)
                for (auto k : s)
                    left.insert(k);
        }

        // bug2(x, left.size());
        res = max(res, (int)left.size());
    }
    cout << res << endl;
}
// void _()
// {
//     int n;
//     cin >> n;
//     map<int, map<int, int>> has;

//     for (int i = 0; i < n; i++)
//     {
//         int cnt;
//         cin >> cnt;
//         set<int> s;
//         for (int i = 0; i < cnt; i++)
//         {
//             int x;
//             cin >> x;
//             s.insert(x);
//         }
//         for (auto x : s)
//             for (auto y : s)
//                 has[x][y]++;
//     }
//     map<int, int> w;
//     for (auto [x, hs] : has)
//     {
//         int c = 0;
//         for (auto [y, cnt] : hs)
//             c += cnt == 1;
//         w[x] = c;
//     }
//     int res = 0;
//     for (auto [x, v] : w)
//         res = max(res, (int)w.size() - v), bug2(x, v);
//     cout << res << endl;
// }

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