TopCoder 250 points 8-SRM 147 DIV 2 176.36/250 70.54%

Problem Statement

 

Julius Caesar used a system of cryptography, now known as Caesar Cipher, which shifted each letter 2 places further through the alphabet (e.g. 'A' shifts to 'C', 'R' shifts to 'T', etc.). At the end of the alphabet we wrap around, that is 'Y' shifts to 'A'.

We can, of course, try shifting by any number. Given an encoded text and a number of places to shift, decode it.

For example, "TOPCODER" shifted by 2 places will be encoded as "VQREQFGT". In other words, if given (quotes for clarity) "VQREQFGT" and 2 as input, you will return "TOPCODER". See example 0 below.

Definition

 Class:CCipherMethod:decodeParameters:String, intReturns:StringMethod signature:String decode(String cipherText, int shift)(be sure your method is public)  

Constraints

-cipherText has between 0 to 50 characters inclusive-each character of cipherText is an uppercase letter 'A'-'Z'-shift is between 0 and 25 inclusive

Examples

0)  

"VQREQFGT"

2

Returns: "TOPCODER"

 1)  

"ABCDEFGHIJKLMNOPQRSTUVWXYZ"

10

Returns: "QRSTUVWXYZABCDEFGHIJKLMNOP"

 2)  

"TOPCODER"

0

Returns: "TOPCODER"

 3)  

"ZWBGLZ"

25

Returns: "AXCHMA"

 4)  

"DBNPCBQ"

1

Returns: "CAMOBAP"

 5)  

"LIPPSASVPH"

4

Returns: "HELLOWORLD"

 

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这道也非常简单，DIV 2里面简单的练手题

public class CCipher {public  String decode(String cipherText, int shift) {int len = cipherText.length();if (len == 0)return "";StringBuilder sb = new StringBuilder();char array[] = cipherText.toCharArray();for (int i = 0; i < len; i++) {int a = array[i] - shift;int b = a + 26;array[i] = array[i] - 'A' >= shift ? (char) a : (char) b;sb.append(array[i]);}return sb.toString();}}