Class assignment

#1

whats wrong in my code

#include
using namespace std;
void class1(int n,string s,int &ans){
if(n==0){

 ans++;
 return;

}
if(n<0){
return;
}
class1(n-1,“a”+s,ans);
int x=s.length();

 if(s[x-1]!='b'){
	 class1(n-1,"b"+s,ans);
 }

}
int main() {
int t;
cin>>t;
while(t–){
int n;
cin>>n;
int cnt=0;
class1(n,"",cnt);
cout<<"#"<<n<<" : "<<cnt<<endl;
}
return 0;
}

#2

Hello @kumarakash121005,

I have implemented the correct logic for you, you can check out here.

#3

why this is wrong
and this code is same as fibonacci

#4

@kumarakash121005,

Your implementation is wrong, for example:

N = 6
1
2
3
4
5
6

Your code’s output:

#1 : 2
#2 : 3
#3 : 5
#4 : 9
#5 : 17
#6 : 33

Clearly, it doesn’t match with the Fibonacci sequence i.e.( 2, 3, 5, 8, 13, 21…)

#5

@kumarakash121005,

Also while printing the answer, you need to print the current test case number instead of n.

cout<<"#"<<testcase_number<<" : "<<answer<<endl;

#6

hum y kaise sure h ki hmari series fibonnaci hi h

#7

@kumarakash121005,

Recursion cases:

  1. If the current element is a then, it is sure that the previous element could be anything either a or b.
    Total ways = f(n-1)

  2. If the current element is b then, it is not possible for the previous element to be b as consecutive elements can’t be equal to b. So, we are left with only 1 case where the previous element = a and, the sequence looks like this:

…ab
Total ways = f(n-2)

F(n) = if current element is a + if current element is b
= f(n-1) + f(n-2)

but this recursion case is similar to Fibonacci.

#8

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