2025.1.5——1200

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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• 2025年第一题...

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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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;
}