Seive function not working

#1
#2

Hey @dishant_rahangdale,

Seive function is correct.
But the actual problem is in your logic.
It seems like if you have not understood the question properly.

  1. You don’t have to only compute the (q)th but the first q(i.e. the value of prime variable in your code) prime numbers.

  2. Your code will work only for q=1.
    It is asked in the code that you have to do the entire logic q times.

Read the question carefully:
In ith iteration, you start picking up the cards in Ai-1th pile from the top one by one and check whether the number written on the card is divisible by the ith prime number. If the number is divisible, you stack that card on pile Bi. Otherwise, you stack that card on pile Ai. After Q iterations, cards can only be on pile B1, B2, B3, . . . BQ, AQ . Output numbers on these cards from top to bottom of each piles in order of B1, B2, B3, . . . BQ, AQ .

Explanation:
Lets understand the question, it will help you to figure out the approach:

N is the no. of cards in A(0)th stack.
Q is the no. of iterations you have to perform.
The third input is N numbers, representing the number of cards present in A(0)th stack.

Let’s understand this with the help of an example:
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):

Pop top element of A(i-1)th (A(0)th) stack i.e. 5
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):

Pop top element of A(i-1)th (A(1)th) stack i.e. 1
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

Note: top element of the stack is popped first.

Hope, this would help.
Give a like if you are satisfied.

#3

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