Coding Problem (Java) __ algorithm

1. Title:

The topic Description assumes that a coded range is 25 letters of a ~ y, from 1-bit to 4-bit encoding, if we sort the encoding in dictionary order, form an array as follows: A, AA, AAA, AAAA, Aaab, AAAC, ..., b, Ba, BAA, b AAA, Baab, Baac ..., yyyw, Yyyx, yyyy one of the index for 0,AA of the index is 1,AAA index is 2, and so on. Write a function, input is any one encoding, output this encoding corresponding to the index. Enter a description:

Enter a string to be encoded, with a string length less than or equal to 100.

Output Description:

Output the index of this encoding

Example 1 input

Baca

Output

16331

2, the topic Analysis:

is actually a little bit of added. This is a little bit of meaning: start with the first analysis, until the end of the string.
Here's a little bit of math, the range expression: [X,y] An expression that represents all the numbers in a range, and includes the number x in the left range
, excluding the number Y in the right range. Then in the following solution, we use one of these expressions to find the number of expressions in a range of
strings.


Here is an example to understand.


Example:
bdcd

First: b
now, let's ask [a,b] how many strings are in this range.
a:1
ax:25
axx:25*25
axxx:25*25*25
The number of strings with a range of [a,b) is: Math.pow (25,3) +math.pow (25,2) + Math.pow (25,1) +1 = 16,276

second digit: D
Now, let's ask the number of strings in this range [B,BD].
b:1

ba:1
bax:25
baxx:25*25
bb:1
bbx:25 bbxx:25*25 bc:1 bcx:25
bcxx:25*25
The number of [B,BD] strings is: 1+3* (Math.pow (25,2) +math.pow (25,1) +1) = 1954


Third: C
now, we ask [BD, BDC) is the number of strings in this range.
bd:1

bda:1
bdax:25
bdb:1
bdbx:25
The number of strings to [Bd,dbc): 1+2* (Math.pow (25,1) +1) = 53

Fourth: D
Now, let's ask the number of strings in the range [BDC,BDCD].
bdc:1

bdca:1
bdcb:1
bdcc:1
bdcd:1
the number of string digits in the range of [BDC,BDCD]: 1+4* (1) = 5

So BDCD is the 18,288th character, because the count of the characters is starting from 0, so the BDCD subscript should be bit 18287

3, Code:

Package schooloffer17;

Import Java.util.Scanner;

/**
 * Created by Caoxiaohong on 17/11/19 10:54.
 * < encoding ></>/public
class Coding {public
    static void Main (string[] args) {
        Scanner scanner= New Scanner (system.in);
        String str;
        while (Scanner.hasnext ()) {
            str=scanner.nextline ();
            int res=0;
            int len=str.length ();
            res+= (Str.charat (0) -97) * (Math.pow (25,3) +math.pow (25,2) +25+1);
            if (len>1)
                res+= (Str.charat (1) -97) * (Math.pow (25,2) +25+1) +1;
            if (len>2)
                res+= (Str.charat (2) -97) * (25+1) +1;
            if (len>3)
                res+= (Str.charat (3) -97) *1+1;
            System.out.println (res);}}