Is there any way I can generate the following series? [closed] - c++

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Consider the following series sk=s(k-1)(!s(k-1))(!s(k-1))s(k-1). Take s1=1221 and !s1=2112, therefore s2=s1(!s1)(!s1)s1, which means s2=1221211221121221.
My current goal is to generate n elements of this series and then to determine the n-th element. I'm sorry if I made the question sound complicated.
I tried making this work but the code just doesn't show the correct answer or it doesn't work at all. I'd hope to see another perspective on this problem.
This is my code:
#include <iostream>
using namespace std;
int main()
{
int n,k=1,a,b,s=1,p=1,i=1;
a=1;b=2;
cin>>n;
while(i<=n){
if(s==1){
if(k==1){
cout<<a;
i++;
k++;
}
else{if(k==2||k==3){
cout<<b;
i++;
k++;
}
if(k==4){
cout<<a<<" ";
i++;
k=1;p++;
if(p==2) {s=2;p=0;}
}
}}
if(i<=n){
if(s==2){
if(k==1){
cout<<b;
i++;
k++;
}
else{if(k==2||k==3){
cout<<a;
i++;
k++;
}
if(k==4){
cout<<b<<" ";
i++;
k=1;p++;
if(p==2) {s=1;p=0;}
}
}
}
}}
return 0;
}

Answer 1

Here is one possible solution for your sequence generator.
#include <iostream>
#include <string>
// generate inverted term s -> !s
std::string invert(const std::string& s) {
const char a = '1';
const char b = '2';
std::string out;
for (int i = 0; i < s.size(); i++) {
// here one can check for invalid input as well, eg 0
out += s[i] == a ? b : a;
}
return out;
}
// sk = s(k - 1)(!s(k - 1))(!s(k - 1))s(k - 1)
std::string nextSeq(const std::string& s) {
std::string s_inv = invert(s);
return s + s_inv + s_inv + s;
}
// generate nth term of the sequence.
std::string nthSeq(const std::string& s, const int n) {
std::string out = s;
for (int i = 1; i < n; i++) {
out = nextSeq(out);
}
return out;
}
int main() {
std::string s1 = "1221";
// this will 2nd term 1221211221121221
std::cout << nthSeq(s1, 2) << std::endl;
return 0;
}