CF356E.Xenia and String Problem
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题目描述
Xenia the coder went to The Olympiad of Informatics and got a string problem. Unfortunately, Xenia isn't fabulous in string algorithms. Help her solve the problem.
String s is a sequence of characters s1s2... s∣s∣ , where record ∣s∣ shows the length of the string.
Substring s[i... j] of string s is string sisi+1... sj .
String s is a Gray string, if it meets the conditions:
- the length of string ∣s∣ is odd;
- character occurs exactly once in the string;
- either ∣s∣=1 , or substrings and are the same and are Gray strings.
For example, strings "abacaba", "xzx", "g" are Gray strings and strings "aaa", "xz", "abaxcbc" are not.
The beauty of string p is the sum of the squares of the lengths of all substrings of string p that are Gray strings. In other words, consider all pairs of values i,j (1<=i<=j<=∣p∣) . If substring p[i... j] is a Gray string, you should add (j−i+1)2 to the beauty.
Xenia has got string t consisting of lowercase English letters. She is allowed to replace at most one letter of the string by any other English letter. The task is to get a string of maximum beauty.
输入格式
The first line contains a non-empty string t (1<=∣t∣<=105) . String t only consists of lowercase English letters.
输出格式
Print the sought maximum beauty value Xenia can get.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
输入输出样例
输入#1
zzz
输出#1
12
输入#2
aba
输出#2
12
输入#3
abacaba
输出#3
83
输入#4
aaaaaa
输出#4
15
说明/提示
In the first test sample the given string can be transformed into string p = "zbz". Such string contains Gray strings as substrings p[1... 1] , p[2... 2] , p[3... 3] и p[1... 3] . In total, the beauty of string p gets equal to 12+12+12+32=12 . You can't obtain a more beautiful string.
In the second test case it is not necessary to perform any operation. The initial string has the maximum possible beauty.