标签：integers 2025.1 16 int cdot 1200 leq ldots color

2025.1.16——1200

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Q1. 1200

You are given \(3\) integers — \(n\), \(x\), \(y\). Let's call the score of a permutation\(^\dagger\) \(p_1, \ldots, p_n\) the following value:

\[(p_{1 \cdot x} + p_{2 \cdot x} + \ldots + p_{\lfloor \frac{n}{x} \rfloor \cdot x}) - (p_{1 \cdot y} + p_{2 \cdot y} + \ldots + p_{\lfloor \frac{n}{y} \rfloor \cdot y}) \]

In other words, the score of a permutation is the sum of \(p_i\) for all indices \(i\) divisible by \(x\), minus the sum of \(p_i\) for all indices \(i\) divisible by \(y\).

You need to find the maximum possible score among all permutations of length \(n\).

For example, if \(n = 7\), \(x = 2\), \(y = 3\), the maximum score is achieved by the permutation \([2,\color{red}{\underline{\color{black}{6}}},\color{blue}{\underline{\color{black}{1}}},\color{red}{\underline{\color{black}{7}}},5,\color{blue}{\underline{\color{red}{\underline{\color{black}{4}}}}},3]\) and is equal to \((6 + 7 + 4) - (1 + 4) = 17 - 5 = 12\).

\(^\dagger\) A permutation of length \(n\) is an array consisting of \(n\) distinct integers from \(1\) to \(n\) in any order. For example, \([2,3,1,5,4]\) is a permutation, but \([1,2,2]\) is not a permutation (the number \(2\) appears twice in the array) and \([1,3,4]\) is also not a permutation (\(n=3\), but the array contains \(4\)).
Input

The only line of each test case description contains \(3\) integers \(n\), \(x\), \(y\) (\(1 \le n \le 10^9\), \(1 \le x, y \le n\)).

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Q2. 1200

You are given two arrays of integers — \(a_1, \ldots, a_n\) of length \(n\), and \(b_1, \ldots, b_m\) of length \(m\). You can choose any element \(b_j\) from array \(b\) (\(1 \leq j \leq m\)), and for all \(1 \leq i \leq n\) perform \(a_i = a_i | b_j\). You can perform any number of such operations.

After all the operations, the value of \(x = a_1 \oplus a_2 \oplus \ldots \oplus a_n\) will be calculated. Find the minimum and maximum values of \(x\) that could be obtained after performing any set of operations.

Above, \(|\) is the bitwise OR operation, and \(\oplus\) is the bitwise XOR operation.
Input

The first line of each test case contains two integers \(n\) and \(m\) (\(1 \leq n, m \leq 2 \cdot 10^5\)) — the sizes of arrays \(a\) and \(b\).

The second line of each test case contains \(n\) integers \(a_1, a_2, \ldots, a_n\) (\(0 \leq a_i \leq 10^9\)) — the array \(a\).

The third line of each test case contains \(m\) integers \(b_1, b_2, \ldots, b_m\) (\(0 \leq b_i \leq 10^9\)) — the array \(b\).

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

• 数学/结论+位运算

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A1

1. 发现结论：除了共有的位置，计算个数分别分配最大值与最小值。
2. 共有的位置是最小公倍数占的位置。

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A2

1. 位运算：保存 \(b\) 数组的每一位，相当于将原异或和值的某一位进行 \(0/1\) 互换。
2. 奇数时可使原值变大，偶数时可使原值减小。

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

A1

#include <bits/stdc++.h>
#define int long long //
#define endl '\n'     // 交互/调试 关
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
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, x, y;
    cin >> n >> x >> y;
    int hasx = n / x, hasy = n / y;
    int hasxy = n / (x * y / __gcd(x, y));
    hasx -= hasxy;
    hasy -= hasxy;
    // bug3(hasx, hasy, hasxy);
    auto get = [](int st, int ed)
    {
        int cnt = ed - st + 1;
        return (st + ed) * cnt / 2;
    };
    cout << get(n - hasx + 1, n) - get(1, hasy) << endl;
}

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A2

#include <bits/stdc++.h>
#define int long long //
#define endl '\n'     // 交互/调试 关
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
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, m;
    cin >> n >> m;
    int _xor = 0;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        _xor ^= x;
    }
    map<int, bool> has_bit;
    for (int i = 0; i < m; i++)
    {
        int x;
        cin >> x;
        for (int j = 0; x >> j; j++)
            if (x >> j & 1) // wa
                has_bit[j] = 1;
    }
    int mx = _xor, mn = _xor;
    if (n & 1)
    {
        for (auto [bit, has] : has_bit)
        {
            if (mx >> bit & 1)
                ;
            else
                mx += 1ll << bit; //, bug(bit);
            // bug(mx);
        }
    }
    else
    {
        for (auto [bit, has] : has_bit)
            if (mn >> bit & 1)
                mn -= 1ll << bit;
    }
    cout << mn << ' ' << mx << endl;
}

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标签：integers,2025.1,16,int,cdot,1200,leq,ldots,color
From： https://www.cnblogs.com/jkkk/p/18677538