2024.12.11 周三

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 周二2024-12-11

19.2024.12.11 周三2024-12-12

20.2024.12.12 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——120001-1642.2025.1.16——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

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2024.12.11 周三

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Q1. 1100

给定一长度为 $n$ 的数组，你需要执行 $k$ 次操作：每次选择一连续子数组(可为空)，将和作为一元素放到到数组的任意位置。问最后数组和的最大值。

Q2. 1100

给你一长度为$2n$的数组$a$，$1$~$n$各出现2次。让你找出两个大小为$2k$集合$l$,$r$,其中$l$属于$a$₁至$a$_n，$r$属于$a$_n+1至$a$_2n。满足$l$异或和等于$r$异或和。

Q3. 1000

Rudolf has an array $a$ of $n$ integers.
In one operation, he can choose an index $i$ ($2\le i\le n-1$) and assign:

• $a_{i-1}=a_{i-1}-1$
• $a_i=a_i-2$
• $a_{i+1}=a_{i+1}-1$

Rudolf can apply this operation any number of times. Any index $i$ can be used zero or more times.

Can he make all the elements of the array equal to zero using this operation?

Q4. 1100

给你一长度为$n$的数组，$x$为0，$i$:1~n,找到大于$x$且为$a$_i倍数的数,更新$x$。

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

A1. 2点

1.第一次操作必然是选择最大连续子数组，然后将其放到最大连续子数组内。

2.以后的操作就可以看作最大和一变二，二变四的过程。

A2. 2点

1.发现$a$₁至$a$_n这一侧单独出现的数字与右侧相同。

2.同时出现$2$次的数异或和为0。可以先使用出现$2$次的数构造，不够再使用单次出现的数构造。

A3. 1点

1.遍历贪心去减，不够减/最后没减为0则NO

A4. 2点

1.快速找到大于$x$的$a$_i的倍数，直接枚举显然不可行。二分当然可以。

2.数学当然ok，$k$$a$_i>x,=>$k$>=$x$/$a$_i,==>k=$x$/$a$_i+1，取整的时候注意一下，答案就是$k$$a$_i。

------------------------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;
}
const int mod = 1e9 + 7;
int qm(int a, int b)
{
    int res = 1;
    while (b)
    {
        if (b & 1)
            res = res * a % mod; // 不要取模时记得取消
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
int inv(int n)
{
    return qm(n, mod - 2);
}

void _()
{
    int n, k;
    cin >> n >> k;
    int pre = 0, min_pre = 0, max_rangesum = 0;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        pre += x;
        max_rangesum = max(max_rangesum, pre - min_pre);
        min_pre = min(min_pre, pre);
    }
    int res = (max_rangesum % mod * qm(2, k)) % mod;
    res = (res + pre % mod - max_rangesum % mod + (int)(1e7) * mod) % mod;
    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, k;
    cin >> n >> k;
    vector<int> l(n + 1);
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        l[x]++;
    }
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
    }
    set<int> db_l, db_r, sgal;
    for (int i = 1; i <= n; i++)
        if (!l[i])
            db_r.insert(i);
        else if (l[i] == 1)
            sgal.insert(i);
        else
            db_l.insert(i);
    vector<int> res_l, res_r;
    auto get = [&](vector<int> &res, set<int> &l)
    {
        for (auto x : l)
            if (res.size() < k << 1)
                res.push_back(x), res.push_back(x);
        for (auto x : sgal)
            if (res.size() < k << 1)
                res.push_back(x);
    };
    get(res_l, db_l);
    get(res_r, db_r);
    for (auto v : res_l)
        cout << v << ' ';
    cout << endl;
    for (auto v : res_r)
        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;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    int f = 1;
    for (int i = 1; i + 2 <= n; i++)
    {
        if (a[i + 1] < a[i] << 1 || a[i + 2] < a[i])
            f = 0;
        a[i + 1] -= a[i] << 1;
        a[i + 2] -= a[i];
    }
    if (a[n - 1] || a[n])
        f = 0;
    cout << (f ? "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 n;
    cin >> n;
    int x;
    cin >> x;
    int res = x;
    for (int i = 2; i <= n; i++)
    {
        cin >> x;
        int k = res / x + 1;
        res = x * k;
    }
    cout << res << endl;
}