Question

I am trying to generate this pattern on 'c' or 'c++' but I am unable to grasp the logic. Any advice?

1,11,21,1211,111221,312211,......

I just want to input 1, and then the rest of the numbers should automatically be generated.

The pattern is simple,
1 --> one 1
hence next number is 11
11--> two 1
hence next number is 21
21 --> one 2 one 1
hence next number is 1211

and so on...

Answer 1

@ganeshie8

Answer 2

I don't exactly follow the pattern.

Answer 3

1 is spoken as one 1 , so next number is 11
11 is spoken as two 1, so next number is 21
21 is spoken as one 2 one 1 , so next number is 1211
ans so on..

Answer 4

#include <string>
#include <sstream>

std::string lookandsay(const std::string &s)
{
std::ostringstream r;

for (unsigned int i = 0; i != s.length(); ) {
unsigned int new_i = s.find_first_not_of(s[i], i+1);
if (new_i == std::string::npos)
new_i = s.length();

r << new_i - i << s[i];
i = new_i;
}
return r.str();
}

#include <iostream>

int main()
{
std::string laf = "1";

std::cout << laf << std::endl;
for (int i = 0; i < 10; i++) {
laf = lookandsay(laf);
std::cout << laf << std::endl;
}

return 0;
}

Answer 5

happy to help :)

Answer 6

can you add comments in between and perhaps a bit more explanation/elaboration ! ?

Answer 7

bc
:)

Answer 8

ok

Answer 9

@INeedHelpPlease? Isnt that from http://rosettacode.org/wiki/Look-and-say_sequence#C.2B.2B lol

Answer 10

omg they copied my code in <1 hour

Answer 11

dafaq! -_-

Answer 12

Wut. Are you sure they copied from you lol

Answer 13

plagiarism?

Answer 14

Maybe. He even said it was copied even though this is the default coding of C++ programming and he was unable to explain properly.

Answer 15

Anyways heres some advice for improving programming.

1. Learn the assembly language properly (if you read it wrong this might make ALOT of mistakes) (if you use a bytecode compiled language (e.g. Java, C#, Python), learn its assembly language as well) and get your compiler to produce assembly output of your sources (or use a disassembler if your compiler doesn't have that option. For gcc, use the -S command-line flag). Read the assembly code and understand it properly. Experiment with different methods of achieving the same thing. Look at which one is fastest and the best one for you (usually the one which produces the least number of instructions, but not always).
2. Learn compiler theory. This helps with #1 :b
3. Learn a language that's completely different to anything else you've ever used. I recommend Haskell and Scheme. You can skip this step if you want to hehe
4. Read the standard/specification for your chosen language(s). Then you'll know all the caveats and nuances of the features of the language, which will help you learn what is faster and also help you write more maintainable code.

Answer 16

Depending on the design problem they address, design patterns can be classified in different categories, of which the main categories are:
Creational Patterns
Structural Patterns
Behavioral Patterns.

You can research the three.

Answer 17

@shubhamsrg ye sab bc h tum meri baat maano :)

Answer 18

ye to chutiye hai hi, tum bhi chori karke kam chutiyap nahi kiye! :3

Answer 19

zubaan sambhal ke,boss is always right

Answer 20

anyways, my problem still remains unsolved, anyone can explain me the logic in the code ?

Answer 21

*

Answer 22

#include<iostream>
#include<string>
using namespace std;

int main(){
int noOfSeq = 0;
cout << "Please input an integer" << endl;
cin >> noOfSeq;
int countingSeq = 0;
string prevSeq = "1";
cout << "The first " << noOfSeq << " numbers in the sequence are \n1" << endl;
while(countingSeq < (noOfSeq-1)){
string nextSeq = "";
for(int i=0; i<prevSeq.length();i++){

int count = 1;
int digitProcessed=0;
char countingElement = prevSeq[digitProcessed];

while(digitProcessed < prevSeq.length()){

while(prevSeq[i+1] == countingElement){
count ++;
i++;
}
i++;

nextSeq += to_string(count);
nextSeq += countingElement;

digitProcessed += count;
countingElement = prevSeq[digitProcessed];
count = 1;
}
}

cout << nextSeq << endl;
prevSeq = nextSeq;
countingSeq++;

}
return 0;
}

Answer 23

This generates first n numbers of the sequence, where n is the input from the user. It's fun to code this! :D