Codefest

Problem 7: 2 to N

Time limit: 2 secs

Print the numbers from 2 upto a given number, x.

Input :

First line of input contains number of test cases, t.
Next t lines contain each a number, x.

Output :

For each input case, print the output on a new line. Each line contains the numbers from 2 to x separated by a space.

Constraints :

t<15
1<x<100

Sample Input :

2
4
5

Sample Output:

2 3 4
2 3 4 5

Scoring:

USE OF FUNCTIONS IN "math.h" IS NOT ALLOWED
C: Number of non-whitespace characters
D: Number of digits
S: Number of single quotes
B: Number of bitwise operators ( ~, |, &, ^, >>, << )
I: Number of Increment(++) and Decrement(--) operators
F: Number of times "scanf" is used
S = 12000/(C + 6000*D + 300*S + 200*B + 200*I + 200*F)

Solution:

Possible approaches to solve this problem would be:

• Run a for/while loop from 2 to n like below:

  
      for(i=2;i<n;i++)
          printf("%d ",i);
      printf("n\n");
  
  

  It requires the use of digits and hence result score is really low.

• By using characters as in 'c'-'a' which is equal to 2, digits can be replaced. Again this approach scores low as use of single quotes have been constrained

• When a main function is defined as main(i,j,k){}, i is initialized to 1.So, to obtain tha value of 2, we need to multiply 'i' by 2 as in just leftshift the bits by 1 as i<<i or by using increment operator as i++ or decrement operator as -(--(- i)), or by simply using i+i. Using bitwise and increment or decrement operators will result in low scores, but the use of i+i will result in a high score

• There is another concept which can be used here to get 1. The function 'scanf' returns the number of arguments entered. Hence, the statement x=scanf("%d",&t) will give x=1. x=2 can be done in the same way as in the 3rd point.

• Another approach to solve the problem would require the use of 'enum' to get the values as enum v{a,b} takes the value of a as 0 and b as 1. The value of 2 can be obtained either by using enum v{a,b,c} or use enum v{a,b} and using 'b' to get the value '2' by any of the approaches in third point.

References: A.8.4 Enumerations, Dennis Ritchie




Code:

Implementation of 5th approach:

main(x,t,z,i)
{
enum v{a,b};
gets(&x);
t = atoi(&x);
while(t)
{
t=t-b;
gets(&x);
z = atoi(&x);
for(i=b+b;i<=z;i+=b)
(i!=z)?printf("%d ",i):printf("%d\n",i);
}
}

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