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Contents
raw puzzle

Original problem

The Goal

Binary with 0 and 1 is good, but binary with only 0, or almost, is even better! Originally, this is a concept designed by Chuck Norris to send so called unary messages.

Write a program that takes an incoming message as input and displays as output the message encoded using Chuck Norris’ method.

Rules

Here is the encoding principle:

  • The input message consists of ASCII characters (7-bit)
  • The encoded output message consists of blocks of 0
  • A block is separated from another block by a space
  • Two consecutive blocks are used to produce a series of same value bits (only1 or0 values):
    - First block: it is always 0 or 00. If it is 0, then the series contains 1, if not, it contains 0
    - Second block: the number of 0 in this block is the number of bits in the series

Example

Let’s take a simple example with a message which consists of only one character: Capital C. C in binary is represented as 1000011, so with Chuck Norris’ technique this gives:

  • 0 0 (the first series consists of only a single 1)
  • 00 0000 (the second series consists of four 0)
  • 0 00 (the third consists of two 1)

So C is coded as: 0 0 00 0000 0 00

 

Second example, we want to encode the message CC (i.e. the 14 bits 10000111000011) :

  • 0 0 (one single 1)
  • 00 0000 (four 0)
  • 0 000 (three 1)
  • 00 0000 (four 0)
  • 0 00 (two 1)

So CC is coded as: 0 0 00 0000 0 000 00 0000 0 00

Game Input

Input
Line 1: the message consisting of N ASCII characters (without carriage return)
Output
The encoded message
Constraints
0 < N < 100

JavaScript Solution

A solution can be found quite easily by walking over the string and for each character decide if a new block must be created. An implementation can then look as follows:

var MESSAGE = readline();

var res = "";
var pre = -1;

for (var i = 0; i < MESSAGE.length; i++) {

    for (var j = 6; j >= 0; j--) {

        var bit = MESSAGE.charCodeAt(i) >> j & 1;
        if (bit !== pre) {

            if (-1 !== pre) {
                res+= " ";
            }
            res+= 1 == bit ? "0 " : "00 ";
            pre = bit;
        }
        res+= "0";
    }
}
print(res);