2024.12.18 周三

合集 - 日常训练(56)

1.2025.2.19——150002-19 2.11.23 周六2024-11-24 3.11.24 周日2024-11-25 4.11.25 周一日常2024-11-26 5.2024.11.26 周二日常2024-11-26 6.2024.11.27 周三2024-11-27 7.2024.11.28周四2024-11-28 8.2024.11.29 周五2024-11-29 9.2024.11.30 周六2024-11-30 10.2024.12.1 周日2024-12-02 11.2024.12.2 周一2024-12-02 12.2024.12.3 周二2024-12-04 13.2024.12.4 周三2024-12-05 14.2024.12.5 周四2024-12-06 15.2024.12.7 周六2024-12-08 16.2024.12.8 周日2024-12-09 17.2024.12.9 周一2024-12-10 18.2024.12.10 周二2024-12-11 19.2024.12.11 周三2024-12-12 20.2024.12.12 周四2024-12-13 21.2024.12.13 周五2024-12-14 22.2024.12.14 周六2024-12-15 23.2024.12.16 周一2024-12-17 24.2024.12.17 周二2024-12-18

25.2024.12.18 周三2024-12-19

26.2024.12.19 周四2024-12-20 27.2024.12.20 周五2024-12-21 28.2024.12.21 周六2024-12-22 29.2024.12.22 周日2024-12-23 30.2024.12.23 周一2024-12-24 31.2024.12.24 周四2024-12-25 32.2024.12.25 周三2024-12-26 33.2024.12.26 周四2024-12-28 34.2024.12.27 周五2024-12-28 35.2024.12.28 周六2024-12-29 36.2024.12.29 周日2024-12-30 37.2024.12.30 周一01-01 38.2025.1.5——120001-14 39.2025.1.12——120001-14 40.2025.1.14——120001-15 41.2025.1.15——120001-16 42.2025.1.16——120001-17 43.2025.1.17——120001-18 44.2025.1.18——130001-19 45.2025.1.19——130001-20 46.2025.1.20——130001-21 47.2025.1.21——130001-22 48.2025.1.22——130001-24 49.2025.1.24——140001-26 50.2025.1.26——140001-28 51.2025.2.8——140002-09 52.2025.2.9——140002-11 53.2025.2.10——140002-11 54.2025.2.14——140002-14 55.2025.2.15——140002-15 56.2025.2.17——140002-17

2024.12.18 周三

Q1. 1000

You have an array of zeros $a_{1}, a_{2}, \dots, a_{n}$ of length $n$ .

You can perform two types of operations on it:

Choose an index $i$ such that $1 \leq i \leq n$ and $a_{i} = 0$ , and assign $1$ to $a_{i}$ ;
Choose a pair of indices $l$ and $r$ such that $1 \leq l \leq r \leq n$ , $a_{l} = 1$ , $a_{r} = 1$ , $a_{l} + \dots + a_{r} \geq ⌈ \frac{r - l + 1}{2} ⌉$ , and assign $1$ to $a_{i}$ for all $l \leq i \leq r$ .

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

Fedya is playing a new game called "The Legend of Link", in which one of the character's abilities is to combine two materials into one weapon. Each material has its own strength, which can be represented by a positive integer $x$ . The strength of the resulting weapon is determined as the sum of the absolute differences of the digits in the decimal representation of the integers at each position.

Formally, let the first material have strength $X = \overset{―}{x_{1} x_{2} \dots x_{n}}$ , and the second material have strength $Y = \overset{―}{y_{1} y_{2} \dots y_{n}}$ . Then the strength of the weapon is calculated as $| x_{1} - y_{1} | + | x_{2} - y_{2} | + \dots + | x_{n} - y_{n} |$ . If the integers have different lengths, then the shorter integer is padded with leading zeros.

Fedya has an unlimited supply of materials with all possible strengths from $L$ to $R$ , inclusive. Help him find the maximum possible strength of the weapon he can obtain.

An integer $C = \overset{―}{c_{1} c_{2} \dots c_{k}}$ is defined as an integer obtained by sequentially writing the digits $c_{1}, c_{2}, \dots, c_{k}$ from left to right, i.e. $10^{k - 1} \cdot c_{1} + 10^{k - 2} \cdot c_{2} + \dots + c_{k}$ .

Q3. 1000

You are given two arrays $a$ and $b$ both of length $n$ .

You will merge $^{†}$ these arrays forming another array $c$ of length $2 \cdot n$ . You have to find the maximum length of a subarray consisting of equal values across all arrays $c$ that could be obtained.

$^{†}$ A merge of two arrays results in an array $c$ composed by successively taking the first element of either array (as long as that array is nonempty) and removing it. After this step, the element is appended to the back of $c$ . We repeat this operation as long as we can (i.e. at least one array is nonempty).

Q4. 1000

LuoTianyi gave an array $b$ of $n \cdot m$ integers. She asks you to construct a table $a$ of size $n \times m$ , filled with these $n \cdot m$ numbers, and each element of the array must be used exactly once. Also she asked you to maximize the following value:

$\sum_{i = 1}^{n} \sum_{j = 1}^{m} (max_{1 \leq x \leq i, 1 \leq y \leq j} a_{x, y} - min_{1 \leq x \leq i, 1 \leq y \leq j} a_{x, y})$

This means that we consider $n \cdot m$ subtables with the upper left corner in $(1, 1)$ and the bottom right corner in $(i, j)$ ( $1 \leq i \leq n$ , $1 \leq j \leq m$ ), for each such subtable calculate the difference of the maximum and minimum elements in it, then sum up all these differences. You should maximize the resulting sum.

Help her find the maximal possible value, you don't need to reconstruct the table itself.

`------------------------独自思考分割线------------------------`

用时：20 20 40(-2) 14 总：1h34min 虽然都不难，但都不是一眼，需要证明/发现。

A1.

贪心构造，首先两端必须有，自左向右贪心找中间点，发现其坐标最大为 $l a s t + 1 << 1$ 。

A2.

模拟数位发现，在等长度情况下，以第一个不同点为分界线，前面每一位最大贡献就是 $b [i] -^{'} 0^{'}$ ，后面每一位最大贡献就是 $9$ 。
本质就是考虑每一位数的取值范围，相同前缀下 $a [i]$ 只能取 $[0, b [i]]$ ，否则可取 $[0, 9]$ 。

A3.

将连续相同的数看成一块，根据构造方案，同一数组的相同数的2块不可能合并，不同数组则一定可以合并。
那答案显而易见，记录每个数所在块的最大值，答案就是每个数在两数组块的和的最大值。
写的时候直接枚举 $a$ 数组去另一数组找另外的数wa2发，扒数据也没找到原因，果然wa了只有思路/代码有问题，这种情况就是没有考虑一个数没有在两个数组都存在的情况。

A4.

贪心构造，将极差最大的几个数放在左上角， $(2, 2)$ 到 $(n, m)$ 一定可以构造出 $m a x - m i n$ ，还剩下第一行和第一列。
考虑将 $m a x 1$ 放 $(1, 1)$ $m i n 1 / m i n 2$ 放 $(2, 1) / (1, 2)$ ，同时最小值也可放左上角，维护最大值、次大值、最小值、次小值。同时考虑行列，设置函数进行4次计算即可。显然没有方案更优。

`------------------------代码分割线------------------------`

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(6);
    int T = 1;
    cin >> T;
    while (T--)
        _();
    return 0;
}

void _()
{
    int n;
    cin >> n;
    int res = 2;
    int st = 4;
    for (; st < n; st = st + 1 << 1)
        res++;
    if (n == 1)
        res = 1;
    cout << res << endl;
}

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(6);
    int T = 1;
    cin >> T;
    while (T--)
        _();
    return 0;
}

void _()
{
    string a, b;
    cin >> a >> b;
    int n = a.size(), m = b.size();
    string t(max(n, m) - min(n, m), '0');
    if (n < m)
        a = t + a;
    else
        b = t + b;
    n = max(n, m);
    int f = 0, res = 0;
    for (int i = 0; i < n; i++)
    {
        if (!f && a[i] == b[i])
            continue;
        res += f ? 9 : abs(b[i] - a[i]);
        if (a[i] - b[i])
            f = 1;
    }
    cout << res << endl;
}

A3.

#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(6);
    int T = 1;
    cin >> T;
    while (T--)
        _();
    return 0;
}

void _()
{
    int n;
    cin >> n;
    vector<int> a(n + 1), b(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
        cin >> b[i];
    map<int, int> la, lb;
    auto get = [&](vector<int> &a, map<int, int> &la)
    {
        for (int i = 1; i <= n; i++)
        {
            int j = i;
            for (; j <= n && a[j] == a[i]; j++)
                ;
            la[a[i]] = max(la[a[i]], j - i);
            i = j - 1;
        }
    };
    get(a, la);
    get(b, lb);
    int res = 0;
    for (int i = 1; i <= n << 1; i++)
        res = max(res, la[i] + lb[i]);
    // for (auto [x, v] : la)
    //     res = max(res, v + lb[x]);
    cout << res << endl;
}

A4.

#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(6);
    int T = 1;
    cin >> T;
    while (T--)
        _();
    return 0;
}

void _()
{
    int n, m;
    cin >> n >> m;
    vector<int> a(n * m);
    for (int &x : a)
        cin >> x;
    sort(a.begin(), a.end());
    int max1 = a.back(), max2 = a[a.size() - 2];
    int min1 = a[0], min2 = a[1];
    int res = (n - 1) * (m - 1) * (max1 - min1);
    auto cal = [&](int a, int b, int c)
    {
        return (n - 1) * abs(b - a) + (m - 1) * abs(c - a);
    };
    vector<int> t{cal(max1, min1, min2), cal(max1, min2, min1), cal(min1, max2, max1), cal(min1, max1, max2)};
    sort(t.rbegin(), t.rend());
    res += t[0];
    cout << res << endl;
}

posted @ 2024-12-19 10:11 Jkke 阅读(5) 评论(0) 编辑收藏举报

刷新页面返回顶部

登录后才能查看或发表评论，立即登录或者逛逛博客园首页

相关博文：

· 2024.12.17 周二

· 2024.12.22 周日

· Codeforces Round 760 (Div. 3) D. Array and Operations（贪心）

· left 1 Codeforces Round 920 (Div. 3)

· Codeforces Round 916 (Div. 3)

阅读排行：
· 10年+ .NET Coder 心语 ── 封装的思维：从隐藏、稳定开始理解其本质意义
· 地球OL攻略 —— 某应届生求职总结
· 周边上新：园子的第一款马克杯温暖上架
· 提示词工程——AI应用必不可少的技术
· Open-Sora 2.0 重磅开源！

公告

昵称： Jkke
园龄： 6个月
粉丝： 0
关注： 3

+加关注

2025年3月

日

一

二

三

四

五

六

Jke

2024.12.18 周三

2024.12.18 周三

Q1. 1000

Q2. 1000

Q3. 1000

Q4. 1000

`------------------------独自思考分割线------------------------`

用时：20 20 40(-2) 14 总：1h34min 虽然都不难，但都不是一眼，需要证明/发现。

A1.

A2.

A3.

A4.

`------------------------代码分割线------------------------`

A1.

A2.

A3.

A4.

公告

搜索

常用链接

我的标签

合集

随笔档案

阅读排行榜