805B - 3-palindrome - CodeForces Solution


constructive algorithms *1000

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Python Code:

x=int(input());p=''
for i in range(1,x+1):
	if i%4==1 or i%4==2:p+='a'
	else: p+='b'
print(p)

C++ Code:

#include <bits/stdc++.h>
using namespace std;
int main() {


int n;
cin>>n;
string r="";
for(int i=0;i<n;i++){
    r+="b";
    i++;
    if(i==n){
        break;
    }
    r+='b';
    i++;
    if(i==n){
        break;
    }
    r+="a";
    i++;
    if(i==n){
        break;
    }
    
    r+="a";
    
}
cout<<r<<"\n";


return 0;
}


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