Waiter


Problem Statement :


You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the  prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in  from top to bottom in answers. In the next iteration, do the same with the values in stack . Once the required number of iterations is complete, store the remaining values in Ai  in answers , again from top to bottom. Return the answers  array.


Function Description

Complete the waiter function in the editor below.

waiter has the following parameters:

int number[n]: the numbers on the plates
int q: the number of iterations
Returns

int[n]: the numbers on the plates after processing
Input Format

The first line contains two space separated integers, n and q.
The next line contains n space separated integers representing the initial pile of plates, i.e., A.


Solution :



title-img 


                            Solution in C :

In C ++ :




#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <climits>
#include <set>
#include <map>
using namespace std;

int m, n;

const int mycount = 10000;
vector<int> prime_results;
vector<int> sieve(int n)
{
    set<int> primes;
    vector<int> vec;

    primes.insert(2);

    for(int i=3; i<=n ; i+=2)
    {
        primes.insert(i);
    }       

    int p=*primes.begin();
    vec.push_back(p);
    primes.erase(p);

    int maxRoot = sqrt(*(primes.rbegin()));

    while(primes.size() > 0)
    {
        if(p > maxRoot)
        {
            while(primes.size() > 0)
            {
                p=*primes.begin();
                vec.push_back(p);
                primes.erase(p);        
            }
            break;
        }

        int i = p*p;  
        int temp = (*(primes.rbegin()));

        while(i<=temp)
        {
            primes.erase(i);
            i += p;
            i += p;
        }

        p=*primes.begin();
        vec.push_back(p);
        primes.erase(p);
    }

    return vec;
}

void prepare() {
	prime_results = sieve(mycount);
}

int a[100005];
vector<int> thearray[mycount];
void process() {
    int i, j, k, q, l;
    
    prepare();
    scanf("%d %d", &n, &q);
    
    for (i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    
    for (i = 0; i < n; i++) {
        for (j = 0; j < q; j++) {
            k = prime_results[j];
            if (a[i] % k == 0) {
                thearray[j].push_back(a[i]);
                break;
            }
        }
        
        if (j == q) {
            thearray[j].push_back(a[i]);
        }
    }
    
    for (i = 0; i < q; i++) {
        if ((i & 1) == 0) {
            for (j = 0; j < thearray[i].size(); j++) {
                printf("%d\n", thearray[i][j]);
            }
        } else {
            for (j = thearray[i].size() - 1; j >= 0;j--) {
                printf("%d\n", thearray[i][j]);
            }
        }
    }
    
    if (q & 1) {
        for (j = 0; j < thearray[i].size(); j++) {
            printf("%d\n", thearray[i][j]);
        }
    } else {
        for (j = thearray[i].size() - 1; j >= 0;j--) {
            printf("%d\n", thearray[i][j]);
        }
    }
    
}

int main() {
    process();
    
    return 0;
}









In Java  :






import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    
    public static void main(String[] args) {
      
        
        Scanner sc = new Scanner(System.in);
        String nq[] = sc.nextLine().split(" ");
        int n = Integer.parseInt(nq[0]);
        int q = Integer.parseInt(nq[1]);
        String nums[] = sc.nextLine().split(" ");
        int pnum = 2;
        int primes[] = new int[q];
        MyStack first = new MyStack();
        MyStack rest = new MyStack();
        MyStack dpst = new MyStack();
        for(int i = 0; i < nums.length; ++i){
            first.push(Integer.parseInt(nums[i]));
        }
        for(int i = 0; i < q; ++i){
            if(first.head == null) break;
            primes[i] = pnum;
            while(first.head != null){
                int number = first.pop();
                if(number%pnum == 0)
                    dpst.push(number);
                else
                    rest.push(number);
            }
            while(dpst.head != null){
                System.out.println(dpst.pop());
            }
            pnum = nextPrimeNumber(pnum, i, primes);
            MyStack temp = rest;
            rest = first;
            first = temp;
        }
        while(first.head != null){
            System.out.println(first.pop());
        }
    }
    static int nextPrimeNumber(int num,int idx,int[] primes){
        if(num == 2) return 3;
        while(true){
            num = num + 2;
            boolean isPrime = true;
            for(int i = 0; i <= idx; ++i){
                if(num%primes[i] == 0){
                    isPrime = false;
                    break;
                }
            }
            if(isPrime) break;
        }
        return num;
    }
}

class MyStack{
    Node head;
    MyStack(){
        head = null;
    }
    void push(int data){
        Node newNode = new Node(data);
        newNode.next = head;
        head = newNode;
    }
    
    int pop(){
        int data = head.data;
        head = head.next;
        return data;
    }
}
class Node{
    int data;
    Node next;
    Node(int d){
        data = d;
        next = null;
    }
}








In C :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define MAX 50005
int top=-1;
int pri[MAX],q,b[MAX],stack[MAX];
void push(int sym){
    stack[++top]=sym;
}
int pop(){
    return stack[top--];
}
int isempty(){
    if(top==-1)
        return 1;
    return 0;
}
void prime(){
    
    pri[0]=2;
    int count,c,i=3;
for ( count = 1 ; count < 1250 ;  )
   {
      for ( c = 2 ; c <= i - 1 ; c++ )
      {
         if ( i%c == 0 )
            break;
      }
      if ( c == i )
      {
         pri[count]=i;
         count++;
      }
      i++;
   }
}
int main() {

     
    prime();
    int n,i,a[MAX];
    scanf("%d %d",&n,&q);
    for(i=0;i<n;i++)
        {
        scanf("%d",&a[i]);
    }
    int k=0;
    for(i=0;i<n;i++)
        {
        if(a[i]%pri[k]==0)
            printf("%d\n",a[i]);
        else
            push(a[i]);
    }
  
    k++;
    int s;
    
    while(k<q)
        {
         int z=0;
         while(!isempty())
             {
              b[z++]=pop();
         }
             for(i=0;i<z;i++)
                 {
                 if(b[i]%pri[k]==0)
                     printf("%d\n",b[i]);
                 else
                     push(b[i]);
             }
        k++;
         }
    int x=0,c[MAX];
    if(k==q)
    {
        while(!isempty())
        {
            c[x++]=pop();
    }
        for(i=x-1;i>=0;i--)
            printf("%d\n",c[i]);
    }
    
    return 0;
}








In Python3 : 





a,b = map(int, input().split(" "))
nums = list(map(int, input().split(" ")))
def prime(x):
    x = x + 1
    primes = []
    for a in range(1, 10000):
        for b in range(2, a):
            if a % b == 0: break
        else:
            primes.append(a)
        if len(primes) == x:
            return primes[1:]

primes = prime(b)
L = []
for i in primes:
    temp_nums = []
    L2 = []
    for j in range(len(nums)):
        num = nums[j]
        if num % i == 0:
            L2 += [num]
        else:
            temp_nums += [num]
    L += L2
    nums = list(reversed(temp_nums))

for i in L:
	print(i)
for i in list(reversed(nums)):
	print(i)








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