18. There are some coins on the table, and Barbara and Jenny take turns to take some. When Barbara takes coins, she can only take 2 or 4; when Jenny takes coins, she can only take 1 or 3. They decide who starts by flipping a coin; the one who takes the last coin wins. Assuming both use their best strategies, when there are 2013 or 2014 coins at the start, ( ) wins.
(A) When there are 2013 coins, Barbara wins; when there are 2014 coins, Jenny wins.
(B) When there are 2013 coins, Jenny wins; when there are 2014 coins, the first player wins.
(C) When there are 2013 coins, Barbara wins; when there are 2014 coins, the second player wins.
(D) When there are 2013 coins, Jenny wins; when there are 2014 coins, Barbara wins.
(E) When there are 2013 coins, the first player wins; when there are 2014 coins, the second player wins.
18. B.

Consider two scenarios:
(1) The number of coins is 2013.

Notice that, according to the game rules, for any five coins, no matter who takes \( n \) coins, the other can take \( 5-n \) coins.

If Jenny goes first, she can take 3 coins, and then, no matter how many \( n \) (where \( n=2 \) or 4) coins Barbara takes, Jenny can take \( 5-n \) coins. Since \( 2010 \equiv 0 \pmod{5} \), Barbara will eventually have no coins left and will lose;

If Barbara goes first, no matter how many \( n \) (where \( n=2 \) or 4) coins Barbara takes, Jenny can take \( 5-n \) coins. Since \( 2010 \equiv 0 \pmod{5} \), there will be 3 coins left. When Barbara takes 2 of them, Jenny can take the remaining 1 coin, and Barbara will have no coins left and will lose.
Therefore, no matter who goes first, Jenny will win.
(2) The number of coins is 2014.
If Jenny goes first, she can take 1 coin, leaving 2013 coins on the table, which is the same as scenario (1), and Jenny will win;

If Barbara goes first, she can take 4 coins, leaving 2010 coins. No matter how many \( m \) (where \( m=1 \) or 3) coins Jenny takes, Barbara can take \( 5-m \) coins. Since \( 2010 \equiv 0 \pmod{5} \), Jenny will eventually have no coins left and will lose. Therefore, Barbara will win.

In summary, with 2013 coins, Jenny wins; with 2014 coins, the first player wins.