Playing with Cards

#1

how to approach this problem ??? im not getting it,
and also tell me if i will require sieve of eratosthenes…!!!

#2

Hey @akb.tech17

It’s a simple brute force approach question. Just keep all numbers in a queue initially. Then keep on putting numbers on B or A according condition till Q iterations. No you don’t need sieve for this.

#3

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#4

hello @akb.tech17 yes you have to prepare prime sieve of 1e5 and then you can perform the operation .
INPUT :
N=5
Q=2
A(0)th stack : 1 2 3 4 5

Processing :
iterate for i =1 to Q;

for i=1(frst iteration):

  1. Pop top element of A(i-1)th (A(0)th) stack i.e. 5
  2. Check, ((top_element i.e. 5)%(ith prime number i.e. 1st prime number, 2)) == 0 ? Push top_element to B(i) i.e. B(1)th stack : Push top_element to A(i)i.e.A(1)th stack;

After repeating the same for all elements of stack A(0):
A(0)=[] i.e. empty as all elements have been popped.
A(1)=[5,3,1] as they are not divisible by 2.
B(1)=[4,2] as they are divisible by 2.

for i=2=Q (second and last iteration):

  1. Pop top element of A(i-1)th (A(1)th) stack i.e. 1
  2. Check, ((top_element i.e. 1)%(ith prime number i.e. 2nd prime number, 3)) == 0 ? Push top_element to B(i) i.e. B(2)th stack : Push top_element to A(i)i.e.A(2)th stack;

After repeating the same for all elements of stack A(1):
A(1)=[] i.e. empty as all elements have been popped.
A(2)=[1,5] as they are not divisible by 3.
B(2)=[3] as they are divisible by 3.

Output:
print elements of B(1),B(2),A(2)

2
4
3
5
1


this is the code for this question try to go through the question and let me know if you have nay doubt .
Happy Learning !!

#5

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