### 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 ;  }

### C Questions

C Questions
C Questions

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment, The underlying machine is an x86 system, Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).
Predict the output or error(s) for the following:

void main()
{
int const * p=5; printf("%d",++(*p));
}
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".
main()
{
char s[ ]="man"; int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
aaaa nnnn
Explanation

### Zoho Interview | Set 1 (Advanced Programming Round)

Third Round: (Advanced Programming Round) Here they asked us to create a “Railway reservation system” and gave us 4 modules. The modules were:
1. Booking
2. Availability checking
3. Cancellation
4. Prepare chart
We were asked to create the modules for representing each data first and to continue with the implementation phase.

My Solution :