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package class_2022_09_3_week;
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// 来自阿里
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// 小红定义一个仅有r、e、d三种字符的字符串中
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// 如果仅有一个长度不小于2的回文子串，那么这个字符串定义为"好串"
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// 给定一个正整数n，输出长度为n的好串有多少个
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// 结果对10^9 + 7取模， 1 <= n <= 10^9
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// 示例：
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// n = 1, 输出0
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// n = 2, 输出3
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// n = 3, 输出18
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// https://www.mashibing.com/question/detail/37485
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public class Code06_RedPalindromeGoodStrings {
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// 暴力方法
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// 为了观察规律
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// 具体方法论，在体系学习班，章节39 : 根据对数器找规律
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public static int num1(int n) {
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char[] path = new char[n];
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return process1(path, 0);
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}
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public static int process1(char[] path, int i) {
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if (i == path.length) {
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int[] dp = getManacherDP(path);
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int cnt = 0;
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for (int p : dp) {
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if (p - 1 > 3) {
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return 0;
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}
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if (p - 1 >= 2) {
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cnt++;
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}
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if (cnt > 1) {
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return 0;
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}
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}
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return cnt == 1 ? 1 : 0;
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} else {
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int ans = 0;
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path[i] = 'r';
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ans += process1(path, i + 1);
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path[i] = 'e';
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ans += process1(path, i + 1);
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path[i] = 'd';
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ans += process1(path, i + 1);
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return ans;
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}
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}
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public static int[] getManacherDP(char[] s) {
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char[] str = manacherString(s);
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int[] pArr = new int[str.length];
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int C = -1;
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int R = -1;
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for (int i = 0; i < str.length; i++) {
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pArr[i] = R > i ? Math.min(pArr[2 * C - i], R - i) : 1;
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while (i + pArr[i] < str.length && i - pArr[i] > -1) {
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if (str[i + pArr[i]] == str[i - pArr[i]])
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pArr[i]++;
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else {
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break;
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}
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}
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if (i + pArr[i] > R) {
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R = i + pArr[i];
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C = i;
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}
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}
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return pArr;
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}
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public static char[] manacherString(char[] s) {
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char[] res = new char[s.length * 2 + 1];
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int index = 0;
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for (int i = 0; i != res.length; i++) {
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res[i] = (i & 1) == 0 ? '#' : s[index++];
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}
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return res;
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}
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// 正式方法
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// 观察规律之后，把规律变成代码
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public static int num2(int n) {
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if (n == 1) {
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return 0;
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}
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if (n == 2) {
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return 3;
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}
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if (n == 3) {
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return 18;
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}
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return 6 * (n + 1);
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}
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public static void main(String[] args) {
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for (int i = 1; i <= 10; i++) {
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System.out.println("长度为" + i + ", 答案:" + num1(i) + "," + num2(i));
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}
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}
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}
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