It turns out that under a plausible assumption about C's strategy, the case where B would need to know C's strategy to decide her own strategy doesn't occur because it's bad for A. Under this assumption, all players have a well-defined optimal strategy independent of the other players' strategies.
Let $S_1$ and $S_2$ be two states brought about by the choices of A and B such that in each state C has an interval $[a_i,b_i]$ of options that will lead to the same payoff for C but not for A and B. The assumption about C's strategy is that if $a_1\le a_2$ and $b_1\le b_2$, then $c_1\le c_2$, where $c_i$ is C's choice in $S_i$. This assumption is satisfied for instance if C always favours A, always favours B, always chooses randomly with uniform distribution in the interval, or has a lucky number that he will pick whenever it's in the interval. It is not satisfied, however, if C's strategy is to "punish" or "deter" B by giving B a smaller payoff the smaller B makes C's payoff.
In Greg's second case, B knows that C will play in the middle, so under the assumption made, she can increase her payoff by moving to the left, as this will not cause C to move to the right. This works until $(b-a)/2=1-b$ (where C would start playing just to the right of B), and thus $b=(a+2)/3$, as David pointed out. The payoff for B in this case is known to be at least $1-b=(1-a)/3$, and B will make this move if it is at least preferable to playing just to the left of A with payoff $a$, that is if $(1-a)/3\ge a$, or $a\le1/4$. Then B will have guaranteed payoff $(1-a)/3$, A will have guaranteed payoff $a$, and C will get half of the rest, $(1-a)/3$, and the remaining $(1-a)/3$ will be split between A and B according to C's choice.
On the other hand, if $1/4\lt a\le1/2$, there are two choices for B. While B can't decide between them without knowing more about C's strategy, A knows that whichever choice B makes it will be worse for A. If B plays just to the left of A, C will play just to the right of A, and A's payoff will be $0$. If B plays on the right, at $(a+2)/3$, then the payoff for C if he plays in the middle or just to the right of B, $(1-a)/3\lt1/4$, would be less than the payoff if he plays just to the left of A, namely $a\gt1/4$. Thus C would play just to the left of A, and the payoff of $(b-a)/2=(1-a)/3\lt1/4$ for A would be less than the guaranteed payoff of $1/4$ for $A$ at $a=1/4$.
Thus, under the stated assumption and without making further assumptions about the choices other players will make among options that don't affect their own payoff, A will play at $1/4$, B will play at $3/4$, and C will play at $c$ anywhere between $1/4$ and $3/4$; the payoff will be $1/4$ to C, $1/8+c/2$ to A and $5/8-c/2$ to B.
Returning to the integer case, since $100$ is divisible by $4$, it seems likely that the optimal strategies (under the stated assumption) are $a=25$, $b=75$ and $25\lt c\lt 75$, but I haven't checked that in detail.
It's interesting that C is worst off; I would have thought that choosing last would be to C's advantage, but it turns out that playing earlier allows one to impose a more favourable spacing.
We can also ask what would happen if C did try to set incentives in his strategy. Of course this only works under the assumption that he announces the strategy and the others believe that he will follow it. There are two different versions of this: We could consider strategies under the assumption that C always directly maximizes his own payoff and only uses options to which he is otherwise indifferent to set incentives; or we could also allow him to promise to choose an otherwise suboptimal result. The second case is more complicated, especially if we allow the same for B; whereas the first case is rather straightforward to solve.
So assume that C always chooses an optimal result for himself and reliably announces how he will decide between options to which he is otherwise indifferent. First assume $a\gt1/4$. In this case B can secure herself a payoff of $1/2$ by playing just to the right of $1-a$, which will cause C to play just to the left of $a$, and there is no incentive that C could offer to B to change that. The payoff to A in this case would be $1/2-a\lt1/4$. Since we already know that A can unconditionally secure a payoff of $1/4$ by playing at $a=1/4$, it follows that A will not play at $a\gt1/4$.
Now assume $a=1/4$. Then B can secure a payoff of at least $1/4$ by playing at $3/4$, so C needs to offer her more than that to cause her to play somewhere else. He can offer to play somewhere between $1/4$ and $1/2$ if B plays at $1$, and threaten to play just to the left of B otherwise. This will lead to a payoff of $3/8$ for C, $(1/4+c)/2$ for A and $(1-c)/2$ for B, where the payoffs for A and B both lie between $1/4$ and $3/8$ and depend on the remaining choice by C.
Now assume $a\lt1/4$. Then B can secure a payoff of $(1-a)/3$ by playing at $(a+2)/3$, and C can get her to play at $1$ by offering to play between $a$ and $1-2(1-a)/3=(1+2a)/3$. The payoff for A will be $(a+c)/2$, and this has to be just more than $1/4$, so C can offer to play at $1/2-a$. This is compatible with offering an incentive to B as long as $1/2-a\lt(1+2a)/3$, that is $a\gt1/10$. Thus, by offering to play at $2/5$, he can get A to play at $1/10$ and B to play at $1$, and the payoffs will be $1/4$ to A, $3/10$ to B and $9/20$ to C. (The incentives can be made arbitrarily small, and I didn't include any $\epsilon$s for them so as not to clutter the results.)
Thus, while C is worst off if he plays a simple strategy such as favouring A, favouring B or choosing randomly among options with equal payoff for himself, he is best off and gets almost half the payoff if he puts the indeterminacy in his options to strategic use.
For the integer problem, this would correspond to A choosing $10$, B choosing $40$ and C choosing $100$. However, in this case the incentives can't be made arbitrarily small, so B might have to allow A to play at $11$; this can't be analysed in detail since the question doesn't say what happens in case of a tie.