Little Elephant and Divisors CodeChef Solution

Problem -Little Elephant and Divisors CodeChef Solution

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CodeChef Solution CodeChef Solution in C++17

#include<bits/stdc++.h>
using namespace std;
#define all(arr) arr.begin(),arr.end()
#define rep(i,s,e) for(int i=s;i<e;i++)
#define lli long long int
#define ll long long
const ll INF=1e18;
const int mod=1e9+7;

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int t;cin>>t;
	while(t--){
		int n;cin>>n;vector<int>arr(n);for(int i=0;i<n;i++){cin>>arr[i];}
		int gg=arr[0];
		for(int i=1;i<n;i++){
			gg=__gcd(gg,arr[i]);
		}
		if(gg==1){
			cout<<-1<<"\n";continue;
		}int ans=gg;
		for(int i=2;i*i<=gg;i++){
			if(gg%i==0){
				ans=i;break;
			}
		}cout<<ans<<"\n";
	}
	return 0;
}

Little Elephant and Divisors CodeChef Solution in C++14

#include<bits/stdc++.h>
using namespace std;
#define all(arr) arr.begin(),arr.end()
#define rep(i,s,e) for(int i=s;i<e;i++)
#define lli long long int
#define ll long long
const ll INF=1e18;
const int mod=1e9+7;

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int t;cin>>t;
	while(t--){
		int n;cin>>n;vector<int>arr(n);for(int i=0;i<n;i++){cin>>arr[i];}
		int gg=arr[0];
		for(int i=1;i<n;i++){
			gg=__gcd(gg,arr[i]);
		}
		if(gg==1){
			cout<<-1<<"\n";continue;
		}int ans=gg;
		for(int i=2;i*i<=gg;i++){
			if(gg%i==0){
				ans=i;break;
			}
		}cout<<ans<<"\n";
	}
	return 0;
}

Little Elephant and Divisors CodeChef Solution in PYTH 3

# cook your dish here
from math import gcd
from math import sqrt
from functools import reduce
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int,input().split()))
    l = reduce(gcd,a)
    if l>1:
        mind = l
        for i in range(2,int(sqrt(l))+1):
            if l%i==0:
                mind = i
                break
        print(mind)
    else: print(-1)

Little Elephant and Divisors CodeChef Solution in C

#include <stdio.h>
int gcd(int a, int b)
{
    if(a < b)
    {
        int temp = a;
        a=b;
        b=temp;
    }
    if(b == 0)
    {
        return a;
    }
    else
    {
        gcd(b, a%b);
    }
}
int main()
{
    int t;
    
    scanf("%d", &t);
    
    while(t--)
    {
        int n, p;
        
        scanf("%d", &n);
        
        int a[n];
        
        for(int i=0; i<n; i++)
        {
            scanf("%d", &a[i]);
        }
        
        p=a[0];
        
        for(int i=1; i<n; i++)
        {
            p = gcd(a[i], p);
        }
        
        if(p == 1)
        {
            printf("-1\n");
            continue;
        }
        int r=0;
        for(int i=2; i*i<=p; i++)
        {
            if(p%i == 0)
            {
                r++;
                printf("%d\n", i);
                break;
            }
        }
        
        if(r == 0)
        {
            printf("%d\n", p);
        }
    }
}

Little Elephant and Divisors CodeChef Solution in JAVA

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		
		while(T-->0){
		    int n = sc.nextInt();
		    int[] arr = new int[n];
		    int max  = 0;
		    for(int i=0;i<n ;i++){
		        arr[i]= sc.nextInt();
		        max=Integer.max(max,arr[i]);
		    }
		    int res = arr[0];
		    for(int i=1;i<n;i++){
		        res = gcd(arr[i],res);
		    }
		    int result = -1;
		    boolean found =false;
		    for(int i=2;i*i<=res;i++){
		        
		            if(res%i==0){
		                result =i;
		                found=true;
		                break;
		            }
		            
		        
		    }
		        if(found){
		            
		    System.out.println(result);
		        }else{
		            if(res<=1){
		                System.out.println(-1);
		            }else{
		                
		            System.out.println(res);
		            }
		        }    
		        
		    
		    
		}
	}
}

Little Elephant and Divisors CodeChef Solution in PYPY 3

from math import gcd,sqrt
from functools import reduce
for _ in range(int(input())):
    n=int(input())
    l1=list(map(int,input().split()))
    a=reduce(gcd,l1)
    if a==1:
        print(-1)
    else:
        for i in range(2,int(sqrt(a))+1):
            if a%i==0:
                print(i)
                break
        else:
            print(a)

Little Elephant and Divisors CodeChef Solution in PYTH

N = 10**3 # value of N required
def primesieve(N):
	n = N+1
	sieve = [True] * (n/2)
	for i in xrange(3,int(n**0.5)+1,2):
		if sieve[i/2]:
			sieve[i*i/2::i] = [False] * ((n-i*i-1)/(2*i)+1)
		# endif
	# endfor i
	return [2] + [2*i+1 for i in xrange(1,n/2) if sieve[i]]
# end fun
primes = primesieve(N)
def gcd(n, m):
	r = n % m
	while r > 0:
		n = m
		m = r
		r = n % m
	#endwhile
	return m
#end fun
t = int(raw_input())
for i in range(t):
	N = int(raw_input())
	st = raw_input().split()
	g = int(st[0])
	k = 1
	while (k < N) and (g > 1):
		n = int(st[k])
		g = gcd(g,n)
		k += 1
	# endwhile
	r = -1
	if g > 1:
		sq = int(g**0.5)
		q = 0
		p = primes[q]
		while (p <= sq) and (g%p > 0):
			q += 1
			p = primes[q]
		# endwhile
		if g%p == 0:
			r = p
		else:
			r = g
		# endif
	# endif
	print r
# endfor i

Little Elephant and Divisors CodeChef Solution in C#

using System;

namespace CC_E_LittleElephantDivisors
{
	class MainClass
	{
		public static void Main(string[] args)
		{
			int terms = int.Parse(Console.ReadLine());
			int[] result = new int[terms];

			for (int i = 0; i < terms; i++)
			{
				int lenght = int.Parse(Console.ReadLine());
				int[] array = Array.ConvertAll(Console.ReadLine().Trim().Split(), Convert.ToInt32);
				result[i] = array[0];

				for (int j = 1; j < lenght; j++)
				{
					result[i] = GCD(result[i], array[j]);
				}

				if (result[i] == 1)
					result[i] = -1;
				else
				{
					for (int j = 2; j <= Math.Sqrt(result[i]); j++)
					{
						if (result[i] % j == 0)
						{
							result[i] = j;
							break;
						}
					}
				}
			}

			foreach (int divisor in result)
				Console.WriteLine(divisor);

			Console.ReadLine();
		}

		public static int GCD(int a, int b) 
		{
			if (b == 0)
				return a;
			return GCD(b, a % b);
		}
	}
}

Little Elephant and Divisors CodeChef Solution in NODEJS

var input = '';

function cacheInput(data) {
	input += data;
}

function callMain() {
	input = input.split(/\s+/).map(Number);
	main();
}

var readNumber = function() {
	var i = -1;
	return function() {
		return input[++i];
	};
}();

function GCD(a, b) {
	var r = a % b;
	while (r != 0) {
		a = b;
		b = r;
		r = a % b;
	}
	return b;
}

function smallestPrimeFactor(number) {
	for (var s = 2; s * s <= number; ++s) {
		if (number % s == 0) {
			return s;
		}
	}
	return number;
}

function main() {
	var T = readNumber();
	var output = '';
	var smallestPrimes = new Array(100001);
	for (var i = 0; i < 100001; ++i) {
		smallestPrimes[i] = smallestPrimeFactor(i);
	}
	for (var t = 0; t < T; ++t) {
		var N = readNumber();
		var gcd = readNumber();
		for (var i = 1; i < N; ++i) {
			gcd = GCD(gcd, readNumber());
		}
		if (gcd == 1) {
			output += '-1\n';
		}
		else {
			output += smallestPrimes[gcd] + '\n';
		}
	}
	process.stdout.write(output);
	process.exit();
}

process.stdin.resume();
process.stdin.setEncoding('utf8');
process.stdin.on('data', cacheInput).on('end', callMain);

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