Question

In the following C program fragment j, k, n and TwoLong_n are integer variables, and A is an array of integers. The variable n is initialized to an integer $\ge$ 3, and TwoLong_n is initialized to the value of
$2\ast \lfloor lo{g}_2\left(n\right)\rfloor$

for (k = 3; k <= n; k++)
A [k] = 0;
for (k = 2; k <= TwoLog_n; k++)
for (j = k + 1; j <= n; j++)
A[j] = A[j] $\parallel$ (j%k);
for (j = 3; j <= n; j++)
if ( !A[j] )
printf("%d", j);

The set of numbers printed by this program fragment is

• A. {m $\mid$ m $\le$ n, ($\exists$i) [m = i${}^2$!]}
• B. {m $\mid$ m $\le$ n, ($\exists$i) [m = i!]}
• C. { }
• D. {m $\mid$ m $\le$ n, m is prime}

Answer 1

The correct option is C { }

1. First for loop fill array with '0' from index 3 to n.

2. Second for loop location where index j is not multiple of k i.e. non multiple index of k and update index of non multiple index of 4 as 1.

Since we are updating one array A as 1 because of we are doing step 2 for every value of 2 to $2\times lo{g}_{2}n$.

So if (!A[j]) will return false for each index of array 4 because we set each index in step 2. So it will print nothing