CF156C Cipher

题目描述

题目翻译: 对于单词(字符串) $s$ ,保证它只由小写字母组成。 设 $ s$ 的长度为 $len$ ,下文单词(字符串)下标从 $1$ 开始. 如果一个词可以通过零次或多次运算转换成另一个词,则我们认为这两个词的意思是一致的。 运算仅仅包含以下两种方式。 - 方式一:指定任意一个位置 $p$ $(1 \le p \le len)$使 $s_p$ 上的字母变成 **字母表上** 的后一个字母(如 a 变成 b ,x 变成 y),而 $s_{p+1}$ 则要变成 **字母表上** 的前一个字母(如 d 变成 c)。 - 方式二:指定任意一个位置 $p$ $(1 \le p \le len)$使 $s_p$ 上的字母变成 **字母表上** 的前一个字母,而 $s_{p+1}$ 则要变成 **字母表上** 的后一个字母。 你需要回答对于输入的单词,一共有多少种与它意思一致的单词。 另外,对于字母 a ,不能将它变成前一个字母(因为它在字母表上没有前一个字母),同理 字母 z 也不能变成后一个字母。 给出多个单词(字符串),你需要分别对它们做出回答。

输入格式

输出格式

说明/提示

Some explanations about the operation: - Note that for each letter, we can clearly define the letter that follows it. Letter "b" alphabetically follows letter "a", letter "c" follows letter "b", ..., "z" follows letter "y". - Preceding letters are defined in the similar manner: letter "y" precedes letter "z", ..., "a" precedes letter "b". - Note that the operation never changes a word's length. In the first sample you can obtain the only other word "ba". In the second sample you cannot obtain any other word, so the correct answer is $ 0 $ . Consider the third sample. One operation can transform word "klmbfxzb" into word "klmcexzb": we should choose $ p=4 $ , and replace the fourth letter with the following one ("b" $ → $ "c"), and the fifth one — with the preceding one ("f" $ → $ "e"). Also, we can obtain many other words from this one. An operation can transform word "ya" only into one other word "xb". Word "ya" coincides in its meaning with words "xb", "wc", "vd", ..., "ay" (overall there are $ 24 $ other words). The word "klmbfxzb has many more variants — there are $ 3320092814 $ other words that coincide with in the meaning. So the answer for the first word equals $ 24 $ and for the second one equals $ 320092793 $ — the number $ 3320092814 $ modulo $ 10^{9}+7 $