2024.12.21 周六

2024.12.21 周六

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Q1. 1000

Lottery “Three Sevens” was held for $m$ days. On day $i$, $n_i$ people with the numbers $a_{i,1},\dots ,a_{i,n_i}$ participated in the lottery.

It is known that in each of the $m$ days, only one winner was selected from the lottery participants. The lottery winner on day $i$ was not allowed to participate in the lottery in the days from $i+1$ to $m$.

Unfortunately, the information about the lottery winners has been lost. You need to find any possible list of lottery winners on days from $1$ to $m$ or determine that no solution exists.

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Q2. 1000

Kristina has a string $s$ of length $n$, consisting only of lowercase and uppercase Latin letters. For each pair of lowercase letter and its matching uppercase letter, Kristina can get $1$ burl. However, pairs of characters cannot overlap, so each character can only be in one pair.

For example, if she has the string $s$ = “aAaaBACacbE”, she can get a burl for the following character pairs:

• $s_1$ = “a” and $s_2$ = “A”
• $s_4$ = “a” and $s_6$ = “A”
• $s_5$ = “B” and $s_{10}$ = “b”
• $s_7$= “C” and $s_9$ = “c”

Kristina wants to get more burles for her string, so she is going to perform no more than $k$ operations on it. In one operation, she can:

• either select the lowercase character $s_i$ ($1\le i\le n$) and make it uppercase.
• or select uppercase character $s_i$ ($1\le i\le n$) and make it lowercase.

For example, when $k$ = 2 and $s$ = “aAaaBACacbE” it can perform one operation: choose $s_3$ = “a” and make it uppercase. Then she will get another pair of $s_3$ = “A” and $s_8$ = “a”

Find maximum number of burles Kristina can get for her string.

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Q3. 1000

You are given a two-dimensional plane, and you need to place $n$ chips on it.

You can place a chip only at a point with integer coordinates. The cost of placing a chip at the point $(x,y)$ is equal to $|x|+|y|$ (where $|a|$ is the absolute value of $a$).

The cost of placing $n$ chips is equal to the maximum among the costs of each chip.

You need to place $n$ chips on the plane in such a way that the Euclidean distance between each pair of chips is strictly greater than $1$, and the cost is the minimum possible.

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

• Q1 Q3有些意思。1000分的题先到这，位运算，构造，数论，三者最难。

A1.

1. 发现制约关系：当前答案的数后面的数组都不能出现。从后向前扫，未标记的数就可以作为答案。
2. 这是必要性，需要证明一下充分性。

A2.

1. 本质就是大小写匹配的次数，剩下的自己匹配自己(需要代价)。
2. 前一天扫描贪心匹配wa了，现在复习也没时间去找bug。先这样了

A3.

1. 被卡住了 还有些疑问，去学vue写报告了，先这样

------------------------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 m;
    cin >> m;
    vector<vector<int>> a(m);
    vector<int> vis(50010);
    for (auto &v : a)
    {
        int n;
        cin >> n;
        while (n--)
        {
            int x;
            cin >> x;
            v.push_back(x);
        }
    }
    vector<int> res;
    for (int i = m - 1; i >= 0; i--)
    {
        auto v = a[i];
        int f = 0;
        for (auto t : v)
        {
            if (!vis[t] && !f)
            {
                f = 1;
                res.push_back(t);
            }
            vis[t] = 1;
        }
    }
    if (res.size() - m)
        res.assign(1, -1);
    reverse(res.begin(), res.end());

    for (auto v : res)
        cout << v << " ";
    cout << 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 _()
{
    int m;
    cin >> m;
    vector<vector<int>> a(m);
    vector<int> vis(50010);
    for (auto &v : a)
    {
        int n;
        cin >> n;
        while (n--)
        {
            int x;
            cin >> x;
            v.push_back(x);
        }
    }
    vector<int> res;
    for (int i = m - 1; i >= 0; i--)
    {
        auto v = a[i];
        int f = 0;
        for (auto t : v)
        {
            if (!vis[t] && !f)
            {
                f = 1;
                res.push_back(t);
            }
            vis[t] = 1;
        }
    }
    if (res.size() - m)
        res.assign(1, -1);
    reverse(res.begin(), res.end());

    for (auto v : res)
        cout << v << " ";
    cout << 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;
    int res = ceil(sqrtl(n)) - 1;
    cout << res << endl;
}
// void _()
// {
//     int n;
//     cin >> n;
//     int qrt_n = sqrt(n);
//     int k = qrt_n / 2 + (qrt_n * qrt_n != n);
//     int res = k - 1 << 1 | 1;
//     if ((k - 1) * (k - 1) << 2 == n - 1)
//         res--;
//     if (n == 1)
//         res = 0;
//     // bug(k);
//     cout << res << endl;
// }

原文地址：https://blog.csdn.net/2302_79354434/article/details/144643611

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