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Consider the following segment of C code: ```c int j, n; j = 1; while (j <= n) j = j * 2; ``` The number of comparisons made in the execution of the loop for any $n>0$ is:

1. $\lfloor \log_2 n \rfloor + 1$
2. $n$
3. $\lfloor \log_2 n \rfloor$
4. $\lfloor \log_2 n \rfloor + 1$

Correct answer: $\lfloor \log_2 n \rfloor + 1$

Solution

Starting from 1, $j$ takes values $1,2,4,\dots$ until it becomes greater than $n$. The number of successful iterations is $\lfloor \log_2 n \rfloor + 1$ for $n>0$, and the condition is checked one extra time after the final iteration, matching the same count in the given options.

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