题解 | 密码截取

#include <iostream>
#include <string>
#include <vector>
#include <unordered_map>
#include <algorithm>

using namespace std;

// 动态规划求最长回文子串的长度
int longestPalindrome(const string& s) {
    int n = s.size();
    if (n == 0) return 0;

    vector<vector<bool>> dp(n, vector<bool>(n, false));
    int max_length = 1;

    // 单个字符总是回文
    for (int i = 0; i < n; ++i) {
        dp[i][i] = true;
    }

    // 检查长度为2的子串
    for (int i = 0; i < n - 1; ++i) {
        if (s[i] == s[i + 1]) {
            dp[i][i + 1] = true;
            max_length = 2;
        }
    }

    // 检查长度大于2的子串
    for (int length = 3; length <= n; ++length) {
        for (int i = 0; i <= n - length; ++i) {
            int j = i + length - 1;
            if (s[i] == s[j] && dp[i + 1][j - 1]) {
                dp[i][j] = true;
                max_length = length;
            }
        }
    }

    return max_length;
}

// 预处理字符串，移除只出现一次的字符
string preprocessString(const string& s) {
    unordered_map<char, int> charCount;
    for (char c : s) {
        charCount[c]++;
    }

    string result;
    for (char c : s) {
        if (charCount[c] > 1) {
            result += c;
        }
    }
    return result;
}

int main() {
    string input;
    getline(cin, input);

    // 预处理字符串
    string processedString = preprocessString(input);

    // 计算最长对称子串的长度
    int maxLength = longestPalindrome(processedString);

    // 输出结果
    cout << maxLength << endl;

    return 0;
}