# Alice and Bob's Silly Game

### Problem Statement :

```Alice and Bob invented the following silly game:

The game starts with an integer, , that's used to build a  of  distinct integers in the inclusive range from  to  (i.e., ).
Alice always plays first, and the two players move in alternating turns.
During each move, the current player chooses a prime number, , from . The player then removes  and all of its multiples from .
The first player to be unable to make a move loses the game.
Alice and Bob play  games. Given the value of  for each game, print the name of the game's winner on a new line. If Alice wins, print Alice; otherwise, print Bob.

Note: Each player always plays optimally, meaning they will not make a move that causes them to lose the game if some better, winning move exists.

Input Format

The first line contains an integer, , denoting the number of games Alice and Bob play.
Each line  of the  subsequent lines contains a single integer, , describing a game.

Output Format

For each game, print the name of the winner on a new line. If Alice wins, print Alice; otherwise, print Bob.```

### Solution :

```                            ```Solution in C :

In  C  :

#include <stdio.h>

int main(void) {

int t;scanf("%d",&t);
while(t--)
{
int i,j;
int n;scanf("%d",&n);
int arr[100005];
for(i=1;i<=100000;i++)
arr[i]=1;

for(i=2;i<=100000;i++)
{
if(arr[i]==1)
{
for(j=2*i;j<=100000;j+=i)
arr[j]=0;
}
}
arr[1]=0;
int c=0;
for(i=1;i<=n;i++)
if(arr[i]==1)
c++;
if(c%2==0)
printf("Bob\n");
else
printf("Alice\n");
}

return 0;
}```
```

```                        ```Solution in C++ :

In  C++ :

#include <bits/stdc++.h>

using namespace std;

int N;
int pr[100001];

int main()
{
scanf("%d", &N);
for(int i=2; i<=100000; i++)
pr[i]=1;
for(int i=2; i<=100000; i++) if(pr[i])
for(int j=i*2; j<=100000; j+=i)
pr[j]=0;
for(int i=3; i<=100000; i++)
pr[i]^=pr[i-1];
while(N--)
{
int x;
scanf("%d", &x);
if(pr[x])
printf("Alice\n");
else
printf("Bob\n");
}
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 {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
ArrayList<Integer> primes = new ArrayList<Integer>();
for (int i = 3; i <= 100000; i+=2) {
int sqrt = (int)Math.sqrt(i);
boolean prime = true;
for (int p : primes) {
if (p > sqrt)
break;
if (i%p==0) {
prime = false;
break;
}
}
if (prime)
}
int[] counts = new int[100001];
int currIndex = 0;
for (int i = 0; i <= 100000; i++) {
if (currIndex < primes.size() && primes.get(currIndex) == i)
currIndex++;
counts[i] = currIndex;
}
int g = sc.nextInt();
for (int i = 0; i < g; i++) {
System.out.println(counts[sc.nextInt()]%2==0?"Bob":"Alice");
}
}
}```
```

```                        ```Solution in Python :

In  Python3 :

#!/bin/python3

import sys

prime_pre = [0 for i in range(100001)]
sieve = [0 for i in range(100001)]
for i in range(2, 100000):
if sieve[i] == 0:
prime_pre[i] = prime_pre[i-1]+1
#primes.append(i)
for j in range(i, 100001, i):
sieve[j] = i
else:
prime_pre[i] = prime_pre[i-1]

g = int(input().strip())
for a0 in range(g):
n = int(input().strip())
print("Bob" if prime_pre[n]%2 == 0 else "Alice")```
```

## 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 <

## 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

## Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

## Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from

## Kitty's Calculations on a Tree

Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a

## Is This a Binary Search Tree?

For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a