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点就是需要发现两个字符串的长度
- 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++;elsebreak;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;
}
