Since I am new to Java, I wanted to see if there were better ways of answering the same question. Are parts of my code redundant or maybe there's an easier way to get the same result?
Problem J4: Big Bang Secrets. The
The encoding algorithm is a Caesar cipher with a shift (S\$S\$) that depends on a parameter (K\$K\$) and its position in the word (P\$P\$, where P = 1\$P = 1\$ for the first letter of each word): S = 3 P +\$S = 3\$, K\$P + K\$. For example, when K = 3\$K = 3\$, ZOOM is encoded as FXAB:
- \$S_1 = 3 \times 1 + 3 = 6\$\$S_1 = 3 \cdot 1 + 3 = 6\$, so
Z→ \$\rightarrow\$F - \$S_2 = 3 \times 2 + 3 = 9\$\$S_2 = 3 \cdot 2 + 3 = 9\$, so
O→ \$\rightarrow\$X - \$S_3 = 3 \times 3 + 3 = 12\$\$S_3 = 3 \cdot 3 + 3 = 12\$, so
O→ \$\rightarrow\$A - \$S_4 = 3 \times 4 + 3 = 15\$\$S_4 = 3 \cdot 4 + 3 = 15\$, so
M→ \$\rightarrow\$B
The challenge is to write a decoder. TheThe first line of input contains K\$K\$ (K < 10\$K \lt 10\$). The second line contains the encoded message, containing up to 20 characters in uppercase.
public class Decoder { public static void main(String[] args) { try { BufferedReader in = new BufferedReader(new InputStreamReader( System.in)); int k = Integer.parseInt(in.readLine()); String word = in.readLine(); char[] cArray = word.toCharArray(); for (int i = 0; i < word.length(); i++) { char out = (char) ((cArray[i]) - (((3 * (i + 1)) + k) % ('Z' - 'A'))); if (out < 'A') { char wrap = (char) (('Z' + 1) - ('A' - out)); System.out.print(wrap); } else { System.out.print(out); } } } catch (IOException e) { System.out.println("Error"); } } } 
