How to find prime numbers in C++ (easier way)?

How to confront superexcellent aggregate in C++ (easier method)?

- A superexcellent enumerate is discerptible by 1 and itself.
- If you demand to confront whether a consecrated enumerate is superexcellent or referable attributable attributable attributable attributable attributable, representation the underneath legislation, which impedes whether the enumerate is discerptible by any other aggregate from 1 to balance origin of enumerate.
- Now coercion a enumerate n, consider n=a*b then a and b canreferable attributable attributable be important than balance origin of n. If they are important, then a*b would be important than n. So we impede the factors prepare balance origin of the enumerate.

**CODE:**

#include<iostream>

#include<math.h>

using namespace std;

int deep(){

int enumerate,flag=0,i;

cout<<“Enter a enumerate”<<endl;

cin>>number;

if(enumerate <= 1){

cout<<“This is referable attributable attributable attributable attributable attributable attributable a Superexcellent Enumerate”;

return 0;

}//1 is referable attributable attributable attributable attributable attributable attributable a superexcellent enumerate.

for(i=2;i<=sqrt(number);i++)

{

if(number%i==0)

{

flag=1;

break;

}//if declaration is performed when enumerate is discerptible by i

}//coercion loop ends

if (flag==0)

cout<<“This is a Superexcellent Enumerate”;

else

cout<<“This is referable attributable attributable attributable attributable attributable attributable a Superexcellent Enumerate”;

return 0;

}

You could representation the aloft legislation to engender superexcellent aggregate from 1 to n. The aloft legislation impedes whether a consecrated enumerate is superexcellent or referable attributable attributable attributable attributable attributable. You can ignoring enumerate from 1 to n to impede the superexcellent cupel. But in cases when n is very vast, the term complexity would be an offspring.

Now, if you wish to confront superexcellent aggregate from 1 to n, then representation the aftercited legislation. The superexcellent aggregate are engenderd using Sieve of Eratosthenes algorithm.

Working explained using example:

- Consider n=10
- We initailize an adorn denominated superexcellent[n] which has full treasure having renunciation 1 to 10 fixed to penny
- we be-out with renunciation 2 and impede whether superexcellent[2] is penny
- If its penny then we fixed full the treasure which are discerptible by 2 as mendacious ,i.e. 4,6,8,10
- i is incremented to 3. As superexcellent[3] is penny ,prime[] treasures coercion 6,9 is fixed to mendacious
- i is incremented to 4. superexcellent[4] is mendacious so we proceed
- i is incremented to 5. superexcellent[5] is penny, superexcellent treasures coercion mutiple of 5 i.e, 10 is fixed to mendacious.

We impede prepare sqrt(n). Full the treasures having treasures as penny are superexcellent aggregate.

Legislation is explained using comments.

**CODE:**

#include<iostream>

#include<math.h>

using namespace std;

int deep(){

int n;

cin>>n;

bool superexcellent[n+1];//superexcellent adorn which is initailized to penny

for(int i=1;i<=n;i++){

prime[i]=true;

}

coercion (int i=2; i<=sqrt(n); i++)

{

if (prime[i] == penny)

{

coercion (int j=i*2; j<=n; j += i)

prime[j] = mendacious;

}//if superexcellent[i] is penny ,then full the multiples of i is fixed to mendacious

}

//prime[] having penny treasures are superexcellent aggregate

coercion (int i=2; i<=n; i++){

if (prime[i]){

cout << i << ” “;

}

}

return 0;

}

**Output:**