Tower Breakers, Revisited!
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[1]=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;
//add primes if necessary
if(p.get(p.size()-1) < n){
for(int i=p.get(p.size()-1)+2; i<=n;i+=2){
if(isPrime(i)) p.add(i);
}
}
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[100];
int [] q = new int[100];
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[0];
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()
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