CODEWITHSUNDEEP
c++algorithmprimes
BECOOL
BECOOL

Reputation: 203

integer constant is too large for ‘long’ type when searching for large primes

I am trying to find prime numbers between 1,000,000,000,000 and 1,000,000,100,000. My code is OK if the number of digits is 4 to 8, but if the digits are more than or equal to 12 i am getting:

  1. warning: integer constant is too large for ‘long’ type
  2. Floating point exception
#include <iostream>
#define num 1000000100000
using namespace std;

int main()
{
  bool prime;
  long double sum = 0;
  for (long long int i=1000000000000; i<=num; i++)
  {
    prime = true;
    for(int j=2; j<=i/2; j++)
    {
      if(i%j == 0) prime = false;
    }
     if(prime) sum+=i;
  }
  cout<<sum<<endl;
}

Upvotes: 1

Views: 804

Answers (2)

Mohit Jain
Mohit Jain

Reputation: 30489

It seems you want to add all prime numbers in range mentioned by you. I would strongly suggest segmented sieve for your purpose.

Or you can do a plain sieving. Take an array of size 100001 to store whether 100000000000 + i is prime or not. (better if you can do with 100001 / 2 bits) to store only odd number 100000000001 + 2 * i and sieve out all the odd multiples in range 3 to 1000001 and add the remaining numbers.

But still if you want to stick with your method, I would make a few suggestion to make it run in some practical time.

#include <iostream>
#define num 1000000100000LL  // make long long
using namespace std;

int main()
{
  bool prime;
  long double sum = 0;
  for (long long int i=1000000000001LL; i<=num; i += 2LL)  // Make long long, loop through only odd numbers as even numbers are not prime except 2
  {
    prime = true;
    for(int j=3; j<=1000001; j += 2) // Loop until sqrt(i) only, loop through odd numbers only
    {
      if(i%j == 0) {
        prime = false;
        break;
      }
    }
     if(prime) {
        cout << i << endl;
        sum+=i;
     }
  }
  cout<<sum<<endl;
}

Upvotes: 1

Sami
Sami

Reputation: 155

Use unsigned long long int i=1000000000000LL instead of long long int i=1000000000000LL.

Also you have to use the j also as unsigned long long int because the i/2 might not fit in the int j.

Upvotes: 1

Related Questions