2025.1.5——1200

2025.1.5——1200

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Q1. 1200

Kevin discovered a binary string \(s\) that starts with 1 in the river at Moonlit River Park and handed it over to you. Your task is to select two non-empty substrings\(^{\text{∗}}\) of \(s\) (which can be overlapped) to maximize the XOR value of these two substrings.

The XOR of two binary strings \(a\) and \(b\) is defined as the result of the \(\oplus\) operation applied to the two numbers obtained by interpreting \(a\) and \(b\) as binary numbers, with the leftmost bit representing the highest value. Here, \(\oplus\) denotes the bitwise XOR operation.

The strings you choose may have leading zeros.

\(^{\text{∗}}\)A string \(a\) is a substring of a string \(b\) if \(a\) can be obtained from \(b\) by the deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

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

• 2025年第一题...

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A1.

1. 1点就是需要发现两个字符串的长度
2. 另一点就是考虑最大值要从高位考虑

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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 _()
{
    string s;
    cin >> s;
    int n = s.size();
    s = ' ' + s;
    int has = 1;
    int l = 1, r = n, sub_l = 1, sub_r = 1;
    string t;
    for (int i = 1; i <= n; i++)
        if (s[i] - s[1])
            has = 0;
    if (has)
    {
        cout << l << ' ' << r << ' ' << sub_l << ' ' << sub_r << endl;
        return;
    }

    int f = 0;

    for (int i = 2; i <= n; i++)
        if (!f)
        {
            if (s[i] == '0')
                t += '1', f = 1;
        }
        else
        {
            t += s[i] == '1' ? '0' : '1';
        }
    int m = t.size();
    int res = 0, real_l = 1;
    sub_r = sub_l + m - 1;
    for (; sub_r <= n; sub_l++, sub_r++)
    {
        int ans = 0;
        for (int i = sub_l, j = i - sub_l; i <= sub_r; i++, j++)
            if (s[i] == t[j])
                ans++;
            else
                break;
        if (ans > res)
        {
            res = ans;
            real_l = sub_l;
        }
    }
    sub_l = real_l;
    sub_r = sub_l + m - 1;
    cout << l << ' ' << r << ' ' << sub_l << ' ' << sub_r << endl;
}

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