2024.12.2 周一

Q1. 1100

给定一个数字(32位以内)，使用1,0,-1构造二进制数位，同时保证不能有相邻的非0数存在。

Q2. 1200

给定2个相同数位的数(<=1e100)，任意操作：交换2数中相同位的数字使两数之积最大。

Q3. 1300

前缀后缀板题

Q4. 1400

给定n,m(<=2e6)。a:1_n,b:1m，问：满足a+b是b*gcd(a,b)倍数的(a,b)对数。

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

A1.

发现只需将原本二进制数位中的连续的1替换：1 1 1 0->-1 0 0 1

A2.

发现操作不改变2数之和，则2数之差越小，积则越大。也可直观想象。

操作就是找到最高位不同的数给a，其余每位最大的数给b。

A3.

A4.

数学

------------------------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> res;
    while (n)
    {
        res.push_back(n & 1);
        n >>= 1;
    }
    res.push_back(0);
    n = res.size();

    for (int i = 0; i < n; i++)
        if (res[i])
        {
            int j = i + 1;
            for (; j < n && res[j]; j++)
                res[i] = -1, res[j] = 0;
            if (j > i + 1) // j位置特判
                res[j] = 1;
            i = j - 1;
        }
    cout << res.size() << endl;
    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 _()
{
    string a, b;
    cin >> a >> b;
    // cout << stoll(a) * stoll(b) << endl;
    int f = 0;
    for (int i = 0; i < a.size(); i++)
        if (!f)
        {
            if (a[i] - b[i])
                f = 1;
            if (a[i] < b[i])
                swap(a[i], b[i]);
        }
        else
        {
            if (a[i] > b[i])
                swap(a[i], b[i]);
        }
    cout << a << endl
         << b << 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, k;
    cin >> n >> k;
    vector<int> a(n + 2), pre(n + 2), repre(n + 2);
    for (int i = 1; i <= n; i++)
        cin >> a[i], pre[i] = pre[i - 1] + a[i];
    for (int i = n; i; i--)
        repre[i] = repre[i + 1] + a[i];
    int l, r;
    for (l = 1; l <= n; l++)
        if (pre[l] > k - k / 2)
            break;
    for (r = n; r; r--)
        if (repre[r] > k >> 1)
            break;
    int res = n - max(0ll, r - l + 1);
    if (pre[n] <= k)
        res = n;
    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;
    int res = 0;
    for (int b = 1; b <= m; b++)
        res += (n + b) / b / b;
    cout << res - 1 << endl;
}