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>();
primes.add(2);
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)
primes.add(i);
}
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())
# your code goes here
print("Bob" if prime_pre[n]%2 == 0 else "Alice")
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