### 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=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;
if(p.get(p.size()-1) < n){
for(int i=p.get(p.size()-1)+2; i<=n;i+=2){
}
}

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;
int [] q = new int;
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;
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()```
```

## 2D Array-DS

Given a 6*6 2D Array, arr: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation: a b c d e f g There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print t

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

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## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

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