Description

Transmitting and memorizing information is a task that requires different coding systems for the best use of the available space. A well known system is that one where a number is associated to a character sequence. It is considered that the Words are made only of small characters of the English alphabet a,b,c, …, z (26 characters). From all these words we consider only those whose letters are in lexigraphical order (each character is smaller than the next character).

The coding system works like this: ? The words are arranged in the increasing order of their length. ? The words with the same length are arranged in lexicographical order (the order from the dictionary). ? We codify these words by their numbering, starting with a, as follows: a - 1 b - 2 … z - 26 ab - 27 … az - 51 bc - 52 … vwxyz - 83681 …

Specify for a given word if it can be codified according to this coding system. For the affirmative case specify its code. Input

The only line contains a word. There are some constraints: ? The word is maximum 10 letters length ? The English alphabet has 26 characters. Output

The output will contain the code of the given word, or 0 if the word can not be codified. Sample Input

bf Sample Output

55

題目大意 符合題意的字符串要滿足位于后面字母比前面大的條件，輸入一個字符串，若不滿足該條件，輸出0，否則輸出按照排序系統該字符串之前有多少個字符串。

解題思路 符合答案的字符串包括兩種： 1、字符串長度小于輸入字符串長度。標記輸入字符串的長度為length，假設字符串為 i （i < length），則滿足條件的值為c[26][i]，可以理解為該情況為從26個字母中選取 i 個，從小到大依次排列。 2、字符串長度等于輸入字符串長度。從左至右依次檢索，對于每一位 i 而言，該位字符的取值范圍應至少比前一位字符大一且小于當前字符，即 ch=[str[i-1]+1,str[i])；又因為該位字符之后剩余字符的取值范圍為 ‘z’-ch，剩余字符長度為 length-i-1，即滿足條件的值為 c[‘z’-ch][length-i-1]。

代碼實現