The strategy is quite simple. If the professor has already failed $k$ students, the student in line should say $99-k$. Then the final student to fail will have provided the correct answer.

There is a strategy using induction. Indeed for $n=2$ first says $1$ and if he passes the second says $1$ as well. Otherwise the second says $0$. Now, let $A_n$ be the strategy for a particular $n$. We will find one for $n+1$ as well. Indeed, the first says $n$. If he fails we can apply on the remaining $n$ people the strategy $A_n$ found above since the goal for the teacher is the same (not to pass all of the remaining ones). If he passes we apply $A_n$ with each answer increased by $1$ (taking into account the first student). Since, the goal for the teacher is still the same, this will work too.

Each student follows the following strategy: For each student $A_i$, Let the number of students who failed so far be $f_i$. Then student $A_i$ should say $99-f_i$. It's easy to see that this works. Suppose the number of students who failed afterall is $F \geq 1$. Then look at the last student who failed; $A'$. $A'$ said $99-(F-1)= 100-F$. His answer is right, so all students pass.

Let $f(i)$ denote the number of passes after the $i$-th student where $i \in \{1,...,100\}$ and let each student guess $f(i-1)+100-i$ note that clearly either the last student to fail guesses the number correctly or there is no such student. $\blacksquare$