2024.12.12 周四

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

2024.12.12 周四

Q1. 1000

You have an array $a$ of $n$ integers.

You can no more than once apply the following operation: select three integers $i$ , $j$ , $x$ ( $1 \leq i \leq j \leq n$ ) and assign all elements of the array with indexes from $i$ to $j$ the value $x$ . The price of this operation depends on the selected indices and is equal to $(j - i + 1)$ burles.
What is the least amount of burles you need to spend to make all the elements of the array equal?

Q2. 1000

You are given a positive integer $n$ .

Find a permutation $^{†}$ $p$ of length $n$ such that there do not exist two distinct indices $i$ and $j$ such that $p_{i}$ divides $p_{j}$ and $p_{i + 1}$ divides $p_{j + 1}$ .

Under the constraints of this problem, it can be proven that at least one $p$ exists.

Q3. 1000

Given an array $a$ of $n$ integers, an array $b$ of $m$ integers, and an even number $k$ .

Your task is to determine whether it is possible to choose exactly $\frac{k}{2}$ elements from both arrays in such a way that among the chosen elements, every integer from $1$ to $k$ is included.

Q4. 1000

A certain number $1 \leq x \leq 10^{9}$ is chosen. You are given two integers $a$ and $b$ , which are the two largest divisors of the number $x$ . At the same time, the condition $1 <= a < b < x$ is satisfied.

For the given numbers $a$ , $b$ , you need to find the value of $x$ .

Q5. 1000

You are given a binary string $s$ .

You can perform two types of operations on $s$ :

delete one character from $s$ . This operation costs $1$ coin;
swap any pair of characters in $s$ . This operation is free (costs $0$ coins).

You can perform these operations any number of times and in any order.

Let's name a string you've got after performing operations above as $t$ . The string $t$ is good if for each $i$ from $1$ to $| t |$ $t_{i} \neq s_{i}$ ( $| t |$ is the length of the string $t$ ). The empty string is always good. Note that you are comparing the resulting string $t$ with the initial string $s$ .

What is the minimum total cost to make the string $t$ good?

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

Q2构造 Q4数论比较有意思

A1.

发现只有3种情况，找边界枚举即可。

A2.

倍数就一定相差至少一倍。
如何构造使相邻的2个数都无法找到倍数，1~n存在一半的数无法找到其倍数，因此考虑存在倍数与不存在倍数的数相邻。如：1 n 2 n-1 3 n-2...

A3.

变量维护两数组选的数的个数，map维护两数组有的数。
遍历1~k：考虑独有的，直接加入，若不能加入/都没有则无解。共有的不需要考虑，因为只要二者选的数次数满足，最后一定可以满足。

A4.

对 $x$ 因式分解，b=x/p₁,a=x/p₁ $o r$ x/p₂，...嗯..是道好题。

A5.

任意交换代表任意安排，贪心从前向后构造，直到不能构造，删除剩下的。

`------------------------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;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    int res = n;
    int l = 1;
    for (; l + 1 <= n && a[l + 1] == a[1]; l++)
        ;
    res = min(res, n - l);
    int r = n;
    for (; r > 1 && a[r - 1] == a[n]; r--)
        ;
    res = min(res, r - 1);
    if (a[1] == a[n])
    {
        if (l >= r)
            res = 0;
        else
            res = min(res, r - l - 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 _()
{
    int n;
    cin >> n;
    int l = 1, r = n;
    int f = 1;
    while (l <= r)
    {
        if (f)
            cout << l << ' ', l++;
        else
            cout << r << ' ', r--;
        f ^= 1;
    }
    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, m, k;
    cin >> n >> m >> k;
    map<int, int> ka, kb;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        ka[x] = 1;
    }
    for (int i = 0; i < m; i++)
    {
        int x;
        cin >> x;
        kb[x] = 1;
    }
    bool ans = 1;
    int cnta = 0, cntb = 0;
    for (int i = 1; i <= k; i++)
        if (ka[i] + kb[i] == 1)
        {
            int f = 0;
            if (ka[i] && cnta < k / 2)
                cnta++, f = 1;
            if (kb[i] && cntb < k / 2)
                f = 1, cntb++;
            if (!f)
                ans = f;
        }
        else if (ka[i] + kb[i] == 0)
            ans = 0;

    cout << (ans ? "YES" : "NO") << endl;
}
// void _()
// {
//     int n, m, k;
//     cin >> n >> m >> k;
//     map<int, int> ka, kb;
//     for (int i = 0; i < n; i++)
//     {
//         int x;
//         cin >> x;
//         ka[x]++;
//     }
//     for (int i = 0; i < m; i++)
//     {
//         int x;
//         cin >> x;
//         kb[x]++;
//     }
//     set<int> K;
//     int cnta = 0, cntb = 0;
//     for (int i = 1; i <= k; i++)
//         if (ka[i] + kb[i] == 1)
//         {
//             if (ka[i] && cnta < k / 2)
//                 K.insert(i), cnta++;
//             if (kb[i] && cntb < k / 2)
//                 K.insert(i), cntb++;
//         }
//     for (int i = 1; i <= k; i++)
//         if (ka[i] + kb[i] > 1)
//         {
//             if (cnta < k / 2)
//             {
//                 if (ka[i])
//                 {
//                     K.insert(i), cnta++;
//                     continue;
//                 }
//             }
//             if (cntb < k / 2)
//             {
//                 if (kb[i])
//                 {
//                     K.insert(i), cntb++;
//                     continue;
//                 }
//             }
//         }
//     cout << (K.size() == k ? "YES" : "NO") << 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 a, b;
    cin >> a >> b;
    auto get = [](int n)
    {
        for (int i = 2; i <= n / i; i++)
            if (n % i == 0)
                return i;
        return n;
    };
    int res = b * b / a;
    if (b % a)
        res = b * get(a);
    cout << res << endl;
}

A5

#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 s;
    cin >> s;
    int n = s.size();
    s = " " + s;
    int cnt0 = 0;
    for (int i = 1; i <= n; i++)
        cnt0 += s[i] == '0';
    int cnt1 = n - cnt0;
    int res = 0;
    // bug2(cnt0, cnt1);
    for (int i = 1; i <= n; i++)
    {
        if (s[i] == '0')
        {
            if (!cnt1)
            {
                res = n - (i - 1);
                break;
            }
            cnt1--;
        }
        else
        {
            if (!cnt0)
            {
                res = n - (i - 1);
                break;
            }
            cnt0--;
        }
    }
    cout << res << endl;
}

posted @ 2024-12-13 12:16 Jkke 阅读(5) 评论(0) 编辑收藏举报

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Jke

2024.12.12 周四

2024.12.12 周四

Q1. 1000

Q2. 1000

Q3. 1000

Q4. 1000

Q5. 1000

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

Q2构造 Q4数论比较有意思

A1.

A2.

A3.

A4.

A5.

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

A1.

A2.

A3.

A4.

A5

公告

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2024.12.12 周四

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

Q2构造 Q4数论 比较有意思

A1.

A2.

A3.

A4.

A5.

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

A1.

A2.

A3.

A4.

A5

公告

搜索

常用链接

合集

随笔档案

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

Q2构造 Q4数论比较有意思

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