CCC '12 J4 - Big Bang Secrets - Olympiads Online Judge

CCC '12 J4 - Big Bang Secrets

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Points: 3 (partial)

Time limit: 2.0s

Memory limit: 256M

Problem type

Canadian Computing Competition: 2012 Stage 1, Junior #4

Sheldon and Leonard are physicists who are fixated on the BIG BANG theory. In order to exchange secret insights, they have devised a code that encodes UPPERCASE words by shifting their letters forward.

Shifting a letter by $S$ $S$ positions means to go forward $S$ $S$ letters in the alphabet. For example, shifting B by $S = 3$ $S = 3$ positions gives E. However, sometimes this makes us go past Z, the last letter of the alphabet. Whenever this happens we wrap around, treating A as the letter that follows Z. For example, shifting Z by $S = 2$ $S = 2$ positions gives B.

Sheldon and Leonard's code depends on a parameter $K$ $K$ and also varies depending on the position of each letter in the word. For the letter at position $P$ $P$ , they use the shift value of $S = 3P + K$ $S = 3 P + K$ .

For example, here is how ZOOM is encoded when $K = 3$ $K = 3$ . The first letter Z has a shift value of $S = 3 \times 1 + 3 = 6$ $S = 3 \times 1 + 3 = 6$ ; it wraps around and becomes the letter F. The second letter, O, has $S = 3 \times 2 + 3 = 9$ $S = 3 \times 2 + 3 = 9$ and becomes X. The last two letters become A and B. So Sheldon sends Leonard the secret message: FXAB

Write a program for Leonard that will decode messages sent by Sheldon.

Input Specification

The input will be two lines. The first line will contain the positive integer $K$ $K$ ( $K < 10$ $K < 10$ ), which is used to compute the shift value. The second line of input will be the word, which will be a sequence of uppercase characters of length at most $20$ $20$ .

Output Specification

The output will be the decoded word of uppercase letters.

Sample Input 1

Copy

3
FXAB

Output for Sample Input 1

Copy

ZOOM

Sample Input 2

Copy

5
JTUSUKG

Output for Sample Input 2

Copy

BIGBANG

Comments

2

lidonghan commented on Dec. 3, 2022, 11:57 a.m. ← edited →

Taking $S \bmod 26$ $S mod 26$ also works.

2

Rhumph commented on Sept. 13, 2022, 2:10 p.m.

Looking for a little help…in the second sample…first letter, s=5*1+5=10. If you move 10 spots to the left of j, you get z. The shift s should be 8. What am I missing? Thanks.

2

shibendu commented on Sept. 13, 2022, 2:18 p.m. ← edited →

Shift is $3P+K$ $3 P + K$ , where position $P$ $P$ of 1st letter 'B' is 1. So shift is $3 \times 1+5=8$ $3 \times 1 + 5 = 8$ . Therefore J becomes B after left shift.

0

Rhumph commented on Sept. 13, 2022, 3:34 p.m.

Thank you

1

mahes0640 commented on Feb. 27, 2021, 2:54 p.m.

How do the sample outputs work? When I put 3 and FXAB, and do this formula 3×1+3, how does it go to Z? My program got LGMQ, because when you add the 6 to F (at pos 6) you get to 12 which is L?

9

Josh commented on Feb. 27, 2021, 3:38 p.m.

The question asks you to decode a message, not encode a message.

-2

100nik0v commented on Nov. 18, 2020, 10:57 p.m. ← edited →

Is it just my inexperience, or do you have to use a separate if else loop for every separate letter?

2

Orion222 commented on May 15, 2020, 11:44 p.m.

bruh make the word decode 10 times bigger

1
Harryg33107 commented on Dec. 5, 2019, 2:15 a.m.
My submission produced an AC result. But after I tested it with the following input

Copy
3 HELLOWORLD

It gave the following output:

Copy
BVZWWBQQH=

How come?

7

hxxr commented on Dec. 7, 2019, 10:13 p.m. ← edited →

It is possible for the ASCII character code after decoding to be less than 39. Simply adding 26 to the result will not always make it into a valid letter—sometimes you will need to add a multiple of 26.