Program: Print first 10 Perfect Number

Program: Find or Print first 10 Perfect Number

Perfect number is a positive number which sum of all positive divisors excluding that number.
For example: 6 is perfect number as divisor of 6 are 1, 2, 3. Sum of it divisor is 1 + 2 + 3 = 6.

Simple way to do is find divisor of each element and check for perfect number. But it is slow as it take more time.
So there is general way to find perfect number as given in wikipedia, perfect number exist only for even number. It is given by 2^p−1(2^p − 1) where p is a prime number 
For example for p = 2 : 2¹(2² − 1) = 6,
similarly for p = 3 : 2²(2³ − 1) = 28 ...

So in this program we are printing up to 10 perfect number.

#include <iostream>
#include <cmath> //for pow function
using namespace std;

bool isPrime(int n) //check prime number 
{
//this snippet is taken from program: Check given number for Prime Number
 bool flag = true;
 if(n<2)
  flag = false;
 else if(n == 2)
  flag = true;

        else if(n%2 ==0)
              flag = false;
 else{
  for(int i=3;i<n;i+=2){
   if(n%i == 0)
   {
    flag = false;
     break;
   }
  }

 }
       return flag; //return boolean value
}

/*
using above formula we made function which return the perfect number according to given prime number.
*/
unsigned long long getPerfectNumber(int n){ 
    return (pow(2,n-1)*(pow(2,n)-1));
}


int main()
{
    //print for prime number = 2
    cout << getPerfectNumber(2)<<endl;
    for(int i =3,j=1;j<10;i+=2){
        if(isPrime(i)){ //check prime if found then will print perfect number
                cout << getPerfectNumber(i)<<endl;    
j++;
}
    }
  return 0; //https://cplspls.blogspot.com

}

OUTPUT:

For those who want to do by simple way by check each number for perfect number then here is program:-

#include <iostream>
using namespace std;

int main() {
unsigned long long sum;
        int n = 4; //number of perfect number to print
for(int i=2,j=0;j<n;i++){//we start from 2 and start checking each number 
sum = 0;
for(int j=1;j<i;j++){ //find divisor
if(i%j == 0)
sum+=j; //add divisor
}
if(sum == i){ //compare sum with number
cout << sum<<endl;
j++;
}
}
return 0;
}

Above program sure look small as compare to above one but time taken to print 10 elements will take a lot of time and memory while first code will give answer within few seconds.

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Program: Print first n Prime Number

Program: Print first n Prime Number

We earlier done the program of how to check Prime Number, in that program we are checking whether the given number is prime or not.

In this program we are taking input n and printing first n prime number.

Prime number is a number which is greater than one and can only divide by 1 or by its own number.

#include <iostream>
using namespace std;

//we already discuss about how to check prime number so we create a function
bool isPrime(int n) 
{
       //this snippet is taken from program: Check given number for Prime Number
bool flag = true;
if(n<2)
flag = false;
else if(n == 2)
flag = true;

        else if(n%2 ==0)
              flag = false;
else{
for(int i=3;i<n;i+=2){
if(n%i == 0)
{
flag = false;
 break;
}
}

}
       return flag; //return boolean value
}

int main()
{
  int n,i;
  cout << "Enter value of n : ";
  cin >> n; //input n
  if(n>1)
  {
   cout << "2 ";
   n--;
  }
/*
we print first prime number i.e. 2 earlier to increase the speed of program as i discussed earlier in my last program that 2 is only even prime number. Rest of prime number are odd. So we have to print 2 earlier to run loop with increment of two. This way it checks less number of number and increase its speed.
Now loop run till n value drop to zero that each value we print will decrement value of n. So we can print exactly n number only. 
In loop we pick number and check whether it is prime or not from function which return true or false. According to result we print the prime number or not.
*/
  for(i=2;n>0;i+=2)
  {
      if(isPrime(i))
      {
          cout << i << " ";
          n--;
      }
  }
  return 0; // https://cplspls.blogspot.com
}



OUTPUT:

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