Happy to see Alice play with someone other than Bob. The answer is $2^{50}$. Eva's strategy is to simply pair up consecutive cells (call it a "block" henceforth) on the rows, such that if Alice moves on one of the cells, then Eva moves on the other. Hence there are $2$ possiblities of a block and hence there are $2^{50}$ possible orientations of the rows. $\square$ Next we describe Alice's strategy. Call a previously filled row to be "dangerous" if it matches the current row in all the filled cells. We claim that it is possible for Alice to halve the number of dangerous rows with each of her moves. Assume that after the $i$-th moves from both the players on the current row, there are $x$ dangerous rows. Note that each dangerous row has $50-i$ cross signs over so many empty cells of the current row. Therefore by pigeonhole, there is a cell in the current row with atmost $\frac { x(50-i)}{100-2i}= \frac x2$, cross signs above it. Alice chooses this cell for her next move. Hence if at any moment, less than $2^{50}$ rows have been filled, the initial number of dangerous rows is $<2^{50}$ and Alice can always bring that down to less than $1$, which means she made a new row. We are done. $\square$

The answer is $2^{50}$. We will replace $100$ by $2n$ and show that in general the solution is $2^n$. We first show that Eva can prevent Alice from getting $2^{50}+1$ or more different rows. She will be pairing up column $2i-1$ with $2i$. Whenever Alice marks one cell she will mark the other one in the pair. Now, no rows will have column $2i-1,2i$ both being a cross. Since there is exactly two choices for the coloring of cells $(2i-1,2i)$, there are totally $2^{50}$ possibility for the row combinations. Now we show that Alice can achieve $2^n$ different row combinations. We will call a state where Alice can no longer create new row pattern by a bad state. We claim that a bad state has at least $2^n$ row combinations. We show this by induction on $n$. The case $n=1$ is trivial. Suppose all smaller case holds and consider the case $n$. Consider a bad state, suppose it has $T$ different row combinations. Let $T_{i,j}$ be the set of combinations in $T$ which has a cross in $i$ and a circle in $j$ Now notice that given a cell $1\leq i\leq n$ there exists some $f(i)\neq i$ such that $|T_{i,f(i)}|\geq 2^{n-1}$, otherwise by inductive hypothesis it will not be a good state and Alice can create a new row combination. Now notice that $f(i)$ is a function on the set $\{1,2,...,n\}$ with no fixed points, and so it contains a cycle $V_1\to V_2\to...\to V_k\to V_1$ where $f(V_i)=V_{i+1}$. Now we have $T_{V_i,V_{i+1}}\geq 2^{n-1}$. Meanwhile notice that a combination $t\in T$ can not belongs to both $T_{V_i,V_{i+1}}$ and $T_{V_{i+1},V_{i+2}}$ otherwise its value on the $V_{i+1}$ cell will both be a cross and a circle. Therefore, each $t$ belongs to at most $\lfloor \frac{k}{2}\rfloor$ of $T_{V_i,V_{i+1}}$, hence $$\lfloor\frac{k}{2}\rfloor|T|\geq\frac{k}{2}T\geq \sum_{i=1}^k|T_{V_i,V_{i+1}}|\geq k2^{n-1}$$which implies $|T|\geq 2^n$ as desired. This completes the proof.

Replace Eva with Bob because Alice and Bob must remain! The answer is $2^{50}$. Replace crosses with $1$'s so the strings are just binary strings. Bob's strategy is to pair the columns $\{1, 100\}, \{2, 99\}, \cdots, \{50, 51\}$ and whenever Alice puts a $1$ in one of them, Bob puts a $0$ in the other. This way, any binary string created has exactly one $1$ in each pair, so there are $2^{50}$ possible binary strings at most. For Alice's strategy, suppose she's created $S < 2^{50}$ binary strings so far. She picks the column with the least $1$'s so far and places a $1$ in it and Bob responds by putting a $0$ somewhere. Now, there are $< 2^{49}$ possibilities that were there so far, again she picks the least common column among them. Repeating this, she can ensure that when there are $2$ cells left, there are $< 2^0 = 1$ of them created already and so creates a new one. So Alice can ensure she creates at least $2^{50}$ binary strings. Therefore, the answer is indeed $2^{50}$, as claimed. $\blacksquare$

Replace Alice with $A$, Eva with Bob, Bob with $B$ and cross with $0$. And let's work on general case, so replace $100$ with $2k$. Claim 1: $A$ can create at most $2^k$ rows. Proof: Divide each row's cells into $k$ pairs such that $2i-1^{th}$and $2i^{th}$ cells are in $P_i$. When $A$ puts $1$ to $P_j$, after her move $B$ puts $0$ to empty cell of $P_j$. So at the end each $P_i$ can give $2$ distinct values: $(1,0)$ or $(0,1)$. So the game will end with at most $2^k$ distinct raws $\square$. Claim 2: $A$ can create at least $2^k$ rows. Proof: I'll use strong induction on $k$. Base case are trivial. Let it be true up to $2n-2$ and I will prove for $2n$. For the first half of the table i.e. for $2^{2n-1}\times 2n$ part, $A$ puts $1$ to each raw in her first moves on each rows. In each row there are $2n-1$ empty cells for $B's$ first move. So there is a cell (call it good cell) that $B$ put $0$ to it in his first move at least $\frac{2^{2n-1}}{2n-1}$. We have $\frac{2^{2n-1}}{2n-1}>2^{n-1}$. So from induction hypotise $A$ can create $2^{n-1}$ distinct rows from these startings. For the second half of the board, $A$ puts $1$ to good cells of each row in her first moves on each rows. Then follows same strategy for these part of table and gets $2^{n-1}$ distinct rows from here also. So she gets at least $2^n$ distinct rows $\square$. From these claims we conclude that there will be $2^k$ different rows in the table at the end of game.

Claim: Eva can guarantee there are at most $2^{50}$ strings. Proof. Consider the following strategy: whenever Alice fills $(x, y)$, then Eva fills $(x + 50, y)$ taken $\pmod{100}$. This operation is always designed, and cuts the possibility space as desired. $\blacksquare$ Claim: Alice can force at least $2^{50}$ rows. Proof. Suppose that there are $k < 2^{50}$ filled rows. Call a row dangerous if its overlap with the currently filling row allows for the possibility of the two rows being the same. We now show that Alice can halve the number of dangerous rows each time. When it's Alice's move, she first chooses a column with more zeros than crosses. Then she places a cross in that column. Then after Bob's move, Alice can disregard the filled columns and the rows that aren't dangerous. Since the total number of crosses in unfilled columns and dangerous rows is the same as for zeros, Alice can then again choose such a column. $\blacksquare$ Remark: I grossly misread this problem as Alice and Bob playing on the entire board at once for around two days until told otherwise. From that effort, I've created a natural follow up question. Quote: Suppose that instead, Alice and Bob can place anywhere on the board, instead of just row by row. Show that there exists an $C \times 100$ board on which Alice can guarenteed at least $2^{50}$ distinct rows for some integer $C$.

Quote: Remark: I grossly misread this problem as Alice and Bob playing on the entire board at once for around two days until told otherwise. i dont feel bad about spending 45 minutes misreading this anymore The answer is $2^{50}$. Eva's strategy: Eva pairs up opposite columns (leftmost with rightmost, etc.). When Alice makes a move, Eva fills in the paired column. There are $2$ distinct ways to fill in paired columns, hence at most $2^{50}$ possible distinct rows. Alice's strategy: I claim that Alice can actually fill in the first $2^{50}$ rows distinctly. Suppose that $N<2^{50}$ rows have already been filled distinctly, and we are about to fill row $N+1$. Since each has an equal amount of zeroes and crosses, we can find some column with at most $N/2$ crosses. Place a cross in this in the new row. After (and before) Eva's move, there at most $N/2$ filled rows which "match" row $N+1$ (i.e. have a cross where this row does, and have a zero where this row does). Among these rows, the remaining columns contain an equal number of crosses and zeroes, hence one of the contains at most $N/4$ crosses. Place a cross in this new row. After repeating this $50$ times, there end up being at most $N/2^{50}<1$ rows which can possibly be the same as row $N+1$, hence row $N+1$ ends up being distinct from all the others. Combining these finishes the problem. $\blacksquare$