Tower Breakers, Revisited!


Problem Statement :


cTwo players (numbered  and ) are playing a game of Tower Breakers! The rules of the game are as follows:

Player  always moves first, and both players always move optimally.
Initially there are  towers of various heights.
The players move in alternating turns. In each turn, a player can choose a tower of height  and reduce its height to , where  and  evenly divides .
If the current player is unable to make any move, they lose the game.
Given the value of  and the respective height values for all towers, can you determine who will win? If the first player wins, print ; otherwise, print .

Input Format

The first line contains an integer, , denoting the number of test cases.
Each of the  subsequent lines defines a test case. Each test case is described over the following two lines:

An integer, , denoting the number of towers.
 space-separated integers, , where each  describes the height of tower .

Output Format

For each test case, print a single integer denoting the winner (i.e., either  1  or  2 ) on a new line.



Solution :


                            Solution in C :

In  C  :






#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    long long int q,t,s,i,j,n,k=0,p,temp;
     scanf("%lld",&t);
    for(q=0;q<t;q++){
        scanf("%lld",&n);
        s=0;
        for(i=0;i<n;i++)
        {scanf("%lld",&j);
         k=0;
         //if(j>1){k=1;}
         temp=j;
         for(p=2;p<=sqrt(temp);p++){
             if(j%p==0){
                 while(j%p==0){
                     if(j<=1){break;}
                     j=j/p;
                     k++;
                 }
             }
         }
         if(j>1){k++;}
        s=s^k;               
        }
        if(s==0){printf("2\n");}
        else printf("1\n");
    }
   
    return 0;
}
                        

                        Solution in C++ :

in  C++  :





#include<stdio.h>
#include<algorithm>

using namespace std;

const int maxi=1e6+5;

int cnt[maxi],b[maxi];
int p,q,ans,n,t,tmp,xs,m;

void solve()
{
  scanf("%d",&n);

   xs=0;
  for (int i=0;i<n;i++)
   {
       scanf("%d",&p);
       xs=xs^cnt[p];
    }

    if (xs==0) printf("2\n"); else printf("1\n");
}

void sito()
{
    cnt[1]=0;

    for (int i=2;i<maxi;i++)
    {
        if (b[i]==0)
        {
          for (int j=i;j<maxi;j+=i)
          {
              m=j;
              b[j]=1;
              while (m%i==0)
              {
                cnt[j]++;
                m=m/i;
              }

          }
        }
    }
}

int main()
{
    scanf("%d",&t);

    sito();

    while (t--)
       solve();

  return 0;
}
                    

                        Solution in Java :

In  Java  :







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

public class Solution {
//Day 2: Tower Breakers revisited
	static ArrayList<Integer> p;

    public static boolean isPrime(int i){//odd primes only
        for(int j=0;j<p.size() && p.get(j)<=Math.sqrt(i);j++){
            if(i%p.get(j)==0) return false;
        }
        return true;    
    }
    
    public static int divNo(int n){
        if(n==1) return 0;
        
        int count = 0;
        boolean prime = true;
        //add primes if necessary
        if(p.get(p.size()-1) < n){ 
            for(int i=p.get(p.size()-1)+2; i<=n;i+=2){
                if(isPrime(i)) p.add(i);
            }
        }
        
        for(int i=0,j=n; j>1 && p.size()>i && p.get(i)<=j;i++){
            while(j%p.get(i)==0) {//p[i] divides j
                j/=p.get(i); count++; prime=false;
            }
        }
        if(!prime) return count;
        else return 1;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        int [] s = new int[100];
        int [] q = new int[100];
        p =  new ArrayList<Integer>(Arrays.asList(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97));
        int i,j,n,m;
        for(int tt =0;tt<t;tt++){
            n = in.nextInt();
            
            for(i=0;i<n;i++){
                s[i] = in.nextInt();
                q[i] = divNo(s[i]);
            }
            int nimsum = q[0];
            for(i = 1;i<n;i++){
                nimsum^=q[i];
            }
            
            if(nimsum>0) System.out.println("1");
            else System.out.println("2");   
        }
    }
}
                    

                        Solution in Python : 
                            
In  Python3 :






import random

from fractions import gcd
from itertools import product

def composite(a, d, n, s):
    if pow(a, d, n) == 1:
        return False
    for i in range(s):
        if pow(a, 2**i * d, n) == n-1:
            return False
    return True

def millerrabin(n):
    if n == 1:
        return False
    if n in [2, 31, 73]:
        return True
    if n % 2 == 0:
        return False
    d = n - 1
    s = 0
    while d % 2 == 0:
        d >>= 1
        s += 1
    for a in [31, 73]:
        if composite(a, d, n, s):
            return False
    return True

def pollard(n):
    if n % 2 == 0:
        return 2
    a = random.randint(-100, 100)
    while a == -2:
        a = random.randint(-100, 100)
    x = random.randint(1, 1000)
    y = x
    d = 1
    while d == 1:
        x = (pow(x, 2, n) + a) % n
        y = (pow(y, 2, n) + a) % n
        y = (pow(y, 2, n) + a) % n
        d = gcd(abs(x - y), n)
        if d == n:
            break
    return d

def rho_factor(n):
    if n == 1:
        return {1: 0}
    divisors = [n]
    factors = []
    while divisors:
        d = divisors.pop()
        if millerrabin(d):
            factors.append(d)
            continue
        new_d = pollard(d)
        if d == new_d:
            divisors.append(d)
            continue
        divisors.extend([new_d, d // new_d])
    factor_n = {}
    for x in set(factors):
        factor_n[x] = factors.count(x)
    return factor_n

def grundy(n):
    return sum(v for _, v in rho_factor(n).items())

def towers(array):
    ans = 0
    while array:
        ans ^= grundy(array.pop())
    return [1, 2][ans == 0]

def main():
    for _ in range(int(input())):
        n, array = input(), [int(x) for x in input().split()]
        print(towers(array))

if __name__ == '__main__':
    main()
                    

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