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

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 :


                            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)
       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'};
    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;
	while (nTest--)
		if (m==1) puts("2");
		else puts((n&1)?"1":"2");

                        Solution in Java :

In  Java :

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) {
                n /= i;
                i = 1;
        return answer;

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

                        Solution in Python : 
In  Python3 :

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

View More Similar Problems

Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →

Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →

Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →

Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →

Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →

Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →