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.
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Definition |
| | Class: | CCipher | | Method: | decode | | Parameters: | String, int | | Returns: | String | | Method signature: | String decode(String cipherText, int shift) | | (be sure your method is public) |
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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) | |
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| 1) | |
| "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
| 10
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| Returns: "QRSTUVWXYZABCDEFGHIJKLMNOP"
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| 2) | |
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| 3) | |
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| 4) | |
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| 5) | |
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This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.
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();
}
}
