A12201. Palindromic Magic
编程题
普及/提高-
知识点
题目描述
After learning some fancy algorithms about palindromes, Chouti found palindromes very interesting, so he wants to challenge you with this problem.
Chouti has got two strings $A$ and $B$ . Since he likes [palindromes](https://en.wikipedia.org/wiki/Palindrome), he would like to pick $a$ as some non-empty palindromic substring of $A$ and $b$ as some non-empty palindromic substring of $B$ . Concatenating them, he will get string $ab$ .
Chouti thinks strings he could get this way are interesting, so he wants to know how many different strings he can get.
Chouti has got two strings $A$ and $B$ . Since he likes [palindromes](https://en.wikipedia.org/wiki/Palindrome), he would like to pick $a$ as some non-empty palindromic substring of $A$ and $b$ as some non-empty palindromic substring of $B$ . Concatenating them, he will get string $ab$ .
Chouti thinks strings he could get this way are interesting, so he wants to know how many different strings he can get.
输入格式
The first line contains a single string $A$ ( $1 \le |A| \le 2 \cdot 10^5$ ).
The second line contains a single string $B$ ( $1 \le |B| \le 2 \cdot 10^5$ ).
Strings $A$ and $B$ contain only lowercase English letters.
The second line contains a single string $B$ ( $1 \le |B| \le 2 \cdot 10^5$ ).
Strings $A$ and $B$ contain only lowercase English letters.
输出格式
The first and only line should contain a single integer — the number of possible strings.
输入输出样例
输入 #1
aa aba
输出 #1
6
输入 #2
aaba abaa
输出 #2
15
说明/提示
In the first example, attainable strings are
- "a" + "a" = "aa",
- "aa" + "a" = "aaa",
- "aa" + "aba" = "aaaba",
- "aa" + "b" = "aab",
- "a" + "aba" = "aaba",
- "a" + "b" = "ab".
In the second example, attainable strings are "aa", "aaa", "aaaa", "aaaba", "aab", "aaba", "ab", "abaa", "abaaa", "abaaba", "abab", "ba", "baa", "baba", "bb".
Notice that though "a"+"aa"="aa"+"a"="aaa", "aaa" will only be counted once.
- "a" + "a" = "aa",
- "aa" + "a" = "aaa",
- "aa" + "aba" = "aaaba",
- "aa" + "b" = "aab",
- "a" + "aba" = "aaba",
- "a" + "b" = "ab".
In the second example, attainable strings are "aa", "aaa", "aaaa", "aaaba", "aab", "aaba", "ab", "abaa", "abaaa", "abaaba", "abab", "ba", "baa", "baba", "bb".
Notice that though "a"+"aa"="aa"+"a"="aaa", "aaa" will only be counted once.