2025.1.16——1200

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——120001-16

42.2025.1.16——120001-17

43.2025.1.17——120001-1844.2025.1.18——130001-1945.2025.1.19——130001-2046.2025.1.20——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

2025.1.16——1200

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Q1. 1200

You are given $3$ integers — $n$, $x$, $y$. Let's call the score of a permutation${}^{†}$ $p_1,\dots ,p_n$ the following value:

$$(p_{1\cdot x}+p_{2\cdot x}+\dots +p_{\lfloor \frac{n}{x}\rfloor \cdot x})-(p_{1\cdot y}+p_{2\cdot y}+\dots +p_{\lfloor \frac{n}{y}\rfloor \cdot y})$$

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 $[2,\underset{―}{6},\underset{―}{1},\underset{―}{7},5,\underset{―}{\underset{―}{4}},3]$ and is equal to $(6+7+4)-(1+4)=17-5=12$.

${}^{†}$ 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 — $a_1,\dots ,a_n$ of length $n$, and $b_1,\dots ,b_m$ of length $m$. You can choose any element $b_j$ from array $b$ ($1\le j\le m$), and for all $1\le i\le n$ perform $a_i=a_i|b_j$. You can perform any number of such operations.

After all the operations, the value of $x=a_1\oplus a_2\oplus \dots \oplus a_n$ 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 $m$ ($1\le n,m\le 2\cdot {10}^5$) — the sizes of arrays $a$ and $b$.

The second line of each test case contains $n$ integers $a_1,a_2,\dots ,a_n$ ($0\le a_i\le {10}^9$) — the array $a$.

The third line of each test case contains $m$ integers $b_1,b_2,\dots ,b_m$ ($0\le b_i\le {10}^9$) — 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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