Time Complexity: Primality

Problem Statement :

```A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. Given p integers, determine the primality of each integer and return Prime or Not prime on a new line.

Note: If possible, try to come up with an   primality algorithm, or see what sort of optimizations you can come up with for an O(n) algorithm. Be sure to check out the Editorial after submitting your code.

Function Description

Complete the primality function in the editor below.

primality has the following parameter(s):

int n: an integer to test for primality

Returns

string: Prime if n is prime, or Not prime
Input Format

The first line contains an integer, p, the number of integers to check for primality.
Each of the p subsequent lines contains an integer, n, the number to test.

Constraints

1  <=   p   <=  30
1   <=   n  <=  2 x 10^9```

Solution :

```                            ```Solution in C :

In    C   :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int p;
int flag;
scanf("%d",&p);
for(int a0 = 0; a0 < p; a0++)
{
flag=0;
int n;
scanf("%d",&n);
if(n==1)
{
printf("Not prime\n");
flag=1;
break;
}
if(flag!=1){
for(int i=2;i<=sqrt(n);i++)
{

if(n%i==0 || (n%(n/i))==0)
{
printf("Not prime\n");
flag=1;
break;
}

}
}
if(flag==0)
{
printf("Prime\n");
}
}
return 0;
}```
```

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

In   C++  :

#include <iostream>
#include <math.h>
using namespace std;

int main(){
int soLanNhapVao, soNhapVao;
int soUoc = 1;
cin >> soLanNhapVao;
for(int i = 0;i < soLanNhapVao;i++){
cin >> soNhapVao;
for(int j = 2;j < sqrt(soNhapVao);j++){
if(soNhapVao%j == 0){
soUoc += 1;
}
else{
}
}
if(soNhapVao == 1){
cout << "Not prime" << endl;
}
else if(soUoc >= 2){
cout << "Not prime" << endl;
soUoc = 1;
}
else{
cout << "Prime" << endl;
soUoc = 1;
}
}
}```
```

```                        ```Solution in Java :

In  Java  :

import java.util.*;

public class Solution {

static boolean isPrime(int n) {
for(int i=2;i<=Math.sqrt(n);i++) {
if(n%i==0) {
return false;
}
}
return true;
}

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int numOfTests = sc.nextInt();
for (int i = 0; i < numOfTests; i++) {
int x = sc.nextInt();
String s;
if (x >= 2 && isPrime(x)) {
s = "Prime";
} else {
s = "Not prime";
}
System.out.println(s);
}
}
}```
```

```                        ```Solution in Python :

In   Python3  :

from math import *
p = int(input().strip())
for a0 in range(p):
n = int(input().strip())
f=5
if(n!=1):
for i in range(2,int(sqrt(n))):
if(n%i==0):
f=0
if(f==0 or n==1):
print("Not prime")
else:
print("Prime")```
```

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