@Darklordx/

# TEMPLATE-34

## No description

Files
• Main.java
Main.java
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```
```//TEMPLATE
import java.util.*;
import java.math.*;
public class Main {
static Scanner scan;
public static void println(Object s){System.out.println(s);}
public static void print(Object s){System.out.print(s);}
public static int nextInt(){return scan.nextInt();}
public static long nextLong(){return scan.nextLong();}

public static int nextIntL(){int i = scan.nextInt(); scan.nextLine(); return i;}
public static long nextLongL(){long i = scan.nextLong(); scan.nextLine(); return i;}

public static String nextLine(){return scan.nextLine();}

static void printArray(int arr[])
{
int n = arr.length;
for (int i=0; i<n; ++i)
System.out.print(arr[i]+" ");
System.out.println();
}

public static void main(String[] args) {scan = new Scanner(System.in);solution();}

//Solution goes below:
public static void solution(){
int n = nextInt();
for(int i=0; i<n;i++){
long a = nextLong();
if(isPrime(sqrt)){
println("YES");
}else{
println("NO");
}
}else{
println("NO");
}
}
}
static boolean binarySearch(ArrayList<Long> arr, int l, int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;

// If the element is present at the
// middle itself
if (arr.get(mid) == x)
return true;

// If element is smaller than mid, then
// it can only be present in left subarray
if (arr.get(mid) > x)
return binarySearch(arr, l, mid - 1, x);

// Else the element can only be present
// in right subarray
return binarySearch(arr, mid + 1, r, x);
}

// We reach here when element is not present
// in array
return false;
}
static ArrayList<Long> allPrimes(){
ArrayList<Long> res = new ArrayList<Long>();
for(long i = 0 ; i<1000000; i++){
if(isPrime((int)i)){
}
}
return(res);
}
static boolean isPrime(int a){
if(a<=1){
return(false);
}
if(a==2||a==3||a==5||a==7){
return(true);
}
if(a%2==0||a%3==0){
return(false);
}
int max = (int)((Math.sqrt((double)a)+1)/6);
for(int i = 1; i<=max; i++){
if(a%(6*i-1)==0){
return(false);
}
if(a%(6*i+1)==0){
return(false);
}
}
return(true);
}

}```