2025.1.12——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——120001-14

39.2025.1.12——120001-14

40.2025.1.14——120001-1541.2025.1.15——120001-1642.2025.1.16——120001-1743.2025.1.17——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

2025.1.12——1200

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Q1. 1200

You are given a sequence $a$, consisting of $n$ integers, where the $i$-th element of the sequence is equal to $a_i$. You are also given two integers $x$ and $y$ ($x\le y$).

A pair of integers $(i,j)$ is considered interesting if the following conditions are met:

• $1<=i<j<=n$;
• if you simultaneously remove the elements at positions $i$ and $j$ from the sequence $a$, the sum of the remaining elements is at least $x$ and at most $y$.

Your task is to determine the number of interesting pairs of integers for the given sequence $a$.
Input

• The first line contains three integers $n,x,y$ ($3\le n\le 2\cdot {10}^5$, $1\le x\le y\le 2\cdot {10}^{14}$);
• The second line contains $n$ integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le {10}^9$).

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Q2. 1200

Vasya has two hobbies — adding permutations${}^{†}$ to arrays and finding the most frequently occurring element. Recently, he found an array $a$ and decided to find out the maximum number of elements equal to the same number in the array $a$ that he can obtain after adding some permutation to the array $a$.

More formally, Vasya must choose exactly one permutation $p_1,p_2,p_3,\dots ,p_n$ of length $n$, and then change the elements of the array $a$ according to the rule $a_i:=a_i+p_i$. After that, Vasya counts how many times each number occurs in the array $a$ and takes the maximum of these values. You need to determine the maximum value he can obtain.

${}^{†}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).
Input
The first line of each test case contains a single integer $n$ ($1\le n\le 2\cdot {10}^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le {10}^9$) — the elements of the array $a$.

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Q3. 1200

Jay managed to create a problem of difficulty $x$ and decided to make it the second problem for Codeforces Round #921.

But Yash fears that this problem will make the contest highly unbalanced, and the coordinator will reject it. So, he decided to break it up into a problemset of $n$ sub-problems such that the difficulties of all the sub-problems are a positive integer and their sum is equal to $x$.

The coordinator, Aleksey, defines the balance of a problemset as the GCD of the difficulties of all sub-problems in the problemset.

Find the maximum balance that Yash can achieve if he chooses the difficulties of the sub-problems optimally.
Input

Each test case contains a single line of input containing two integers $x$ ($1\le x\le {10}^8$) and $n$ ($1\le n\le x$).

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

• 观察+数论

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A1.

1. 优化：排序后对于每个数二分找两侧边界。
2. 有点细节

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A2.

1. 发现性质：最后选择的数上下界差小于 $n$。
2. 排序后二分找到最多的数。

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A3.

1. 从答案 $v$ 开始考虑，每个数必然是 $v$ 的倍数，$x$ 亦为 $v$ 的倍数，则 $v$ 必然为 $x$ 的因子，对于每个因子若能够分为至少 $n$ 份，则可以作为答案。

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------------------------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, x, y;
    cin >> n >> x >> y;
    vector<int> a(n + 2);
    int sum = 0;
    for (int i = 1; i <= n; i++)
        cin >> a[i], sum += a[i];
    sort(a.begin() + 1, a.end() - 1, [](int &e1, int &e2)
         { return e1 > e2; });
    int res = 0;
    for (int i = 1; i <= n; i++)
    {
        int s = sum - a[i];
        int l = 0, r = n + 1;
        while (r - l - 1)
        {
            int mid = l + r >> 1;
            if (s - a[mid] >= x)
                r = mid;
            else
                l = mid;
        }
        int d = r;
        l = 0, r = n + 1;
        while (r - l - 1)
        {
            int mid = l + r >> 1;
            if (s - a[mid] <= y)
                l = mid;
            else
                r = mid;
        }
        int u = l;
        // bug2(d, u);
        if (d <= u)
            res += u - d + 1 - (i >= d && i <= u);
    }
    cout << res / 2 << 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(6);
    int T = 1;
    cin >> T;
    while (T--)
        _();
    return 0;
}

void _()
{
    int n;
    cin >> n;
    map<int, int> has;
    for (int i = 1; i <= n; i++)
    {
        int x;
        cin >> x;
        has[x] = 1;
    }
    vector<int> a(1, 0);
    for (auto [x, c] : has)
        a.push_back(x);
    int _n = has.size();
    int res = 1;
    for (int i = 1; i <= _n; i++)
    {
        int l = i, r = _n + 1;
        while (r - l - 1)
        {
            int mid = l + r >> 1;
            if (a[mid] - a[i] < n)
                l = mid;
            else
                r = mid;
            res = max(res, l - i + 1);
        }
    }
    // for (auto v : a)
    //     cout << v << ' ';
    // cout << endl;
    cout << res << endl;
}

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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 x, n;
    cin >> x >> n;
    int res = 1;
    vector<int> t;
    for (int i = 2; i <= x / i; i++)
        if (x % i == 0)
        {
            t.push_back(i);
            if (x / i - i)
                t.push_back(x / i);
        }
    t.push_back(x);
    for (auto v : t)
        if (x / v >= n)
            res = max(res, v);
    cout << res << endl;
}

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