Tower Breakers


Problem Statement :


Two players are playing a game of Tower Breakers! Player  always moves first, and both players always play optimally.The rules of the game are as follows:

Initially there are  towers.
Each tower is of height .
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 a move, they lose the game.
Given the values of  and , determine which player will win. If the first player wins, return . Otherwise, return .

Function Description

Complete the towerBreakers function in the editor below.

towerBreakers has the following paramter(s):

int n: the number of towers
int m: the height of each tower
Returns

int: the winner of the game


Input Format

The first line contains a single integer , the number of test cases.
Each of the next  lines describes a test case in the form of  space-separated integers,  and .



Solution :



title-img


                            Solution in C :

In  C  :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
long int cnt(long int a,int k)
{
   if(k*k>a)
       return a!=1;
   return a%k?cnt(a,k+1):1+cnt(a/k,k+1);
}
int main() {


    int t;
    long int n,m,s;
    char a[2]={'2','1'};
    scanf("%d",&t);
    while(t--)
    {
      s=0;
      scanf("%ld%ld",&n,&m);
      while(m%4==0)
          m/=2;
      if(n%2==0)
         s=0;
      else
         s^=cnt(m,2);
      printf("%c\n",a[!!s]);
    }
    return 0;
}
                        


                        Solution in C++ :

In  C++  :







#include <bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

int main()
{
	long nTest,n,m;
	scanf("%ld",&nTest);
	while (nTest--)
	{
		scanf("%ld%ld",&n,&m);
		if (m==1) puts("2");
		else puts((n&1)?"1":"2");
	}
}
                    


                        Solution in Java :

In  Java :






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

public class Solution {
    
    private static int numPrimeFactors(int n) {
        int answer = 0;
        for (int i=2; i<=n; i++) {
            if (n%i == 0) {
                answer++;
                n /= i;
                i = 1;
            }
        }
        return answer;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int t=0; t<T; t++) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            if (m == 1) {
                System.out.println(2);
                continue;
            }
            if (n%2 == 0) {
                System.out.println(2);
            } else {
                System.out.println(1);
            }
            
        }
        
    }
}
                    


                        Solution in Python : 
                            
In  Python3 :







test = int(input())
for _ in range(test):
    n,m = map(int,input().strip().split())
    if m==1:
        print(2)
    else:
        if n%2==1:
            print(1)
        else:
            print(2)
                    


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