% 计算列表中原子的数目
count_atoms([], 0).
count_atoms([H|T], N) :-
    atomic(H),
    count_atoms(T, N1),
    N is N1 + 1.
count_atoms([H|T], N) :-
    not(atomic(H)),
    count_atoms(T, N).

% 计算匹配的原子总数
match_atoms(_, [], 0).
match_atoms(A, [H|T], N) :-
    A == H,
    match_atoms(A, T, N1),
    N is N1 + 1.
match_atoms(A, [H|T], N) :-
    A \== H,
    match_atoms(A, T, N).

% 计算列表中所有原子的数量
all_atoms_count([], 0).
all_atoms_count([H|T], N) :-
    atomic(H),
    all_atoms_count(T, N1),
    N is N1 + 1.
all_atoms_count([H|T], N) :-
    not(atomic(H)),
    all_atoms_count(T, N).

% Write a predicate called rlen(X, N) to be true when N counts the total number of occurrences of atoms in the list X
rlen([], 0).
rlen([H|T], N) :-
    % First calculate atomic count
    all_atoms_count([H|T], N1),
    % Then iterate through all atoms and match each one individually
    rlen_helper([H|T], N1, N).

rlen_helper([], _, 0).
rlen_helper([H|T], M, N) :-
    atomic(H),
    match_atoms(H, [H|T], K),
    M1 is M - K,
    rlen_helper(T, M1, N1),
    N is K + N1.
rlen_helper([H|T], M, N) :-
    not(atomic(H)),
    rlen_helper(T, M, N).

这里我们使用了三个辅助谓词来实现此问题。 总原子计数器（all_atoms_count），匹配原子计数器（