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Sieve of Eratosthenes ( to List out the prime number upto n )

To find all the prime numbers less than or equal to 30, proceed as follows.
First generate a list of integers from 2 to 30:
 2  3  4  5  6  7  8  9  10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
First number in the list is 2; cross out every 2nd number in the list after it by counting up from 2 in increments of 2 (these will be all the multiples of 2 in the list):
 2  3  4  5  6  7  8  9  10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Next number in the list after 2 is 3; cross out every 3rd number in the list after it by counting up from 3 in increments of 3 (these will be all the multiples of 3 in the list):
 2  3  4  5  6  7  8  9  10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Next number not yet crossed out in the list after 3 is 5; cross out every 5th number in the list after it by counting up from 5 in increments of 5 (i.e. all the multiples of 5):
 2  3  4  5  6  7  8  9  10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Next number not yet crossed out in the list after 5 is 7; the next step would be to cross out every 7th number in the list after 7, but they are all already crossed out at this point, as these numbers (14, 21, 28) are also multiples of smaller primes because 7*7 is greater than 30. The numbers left not crossed out in the list at this point are all the prime numbers below 30:
 2  3     5     7           11    13          17    19          23                29
int main ( )  {      int n,p ;      cin >> n ;      int * arr ;       arr = new int [ n + 1 ] ;         for ( int i = 1 ; i <= n ; i ++ )      {        arr [ i ] = 0 ;      }      for ( int i = 2 ; i <= n ; i ++ ) {          if ( arr [ i ] == 0 )          {   p = i ;              for ( int j = 2 ; p * j <= n ; j ++ )              {                    arr [ p * j ] = 1 ;              }          }      }      for ( int i = 2 ; i <= n ; i ++ ) { if ( arr [ i ] == 0 ) cout << i << endl ; }      delete [ ] arr ;      return 0 ;  }

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Hackerrank > SQL > Basic Select

Select
01-Select All
Given a City table, whose fields are described as +-------------+----------+ | Field       | Type     | +-------------+----------+ | ID          | int(11)  | | Name        | char(35) | | CountryCode | char(3)  | | District    | char(20) | | Population  | int(11)  | +-------------+----------+
write a query that will fetch all columns for every row in the table.

My Solution
SELECT*FROM city;
---------------------------------------------------------------------------------
02-Select by ID
Given a City table, whose fields are described as