我是OCaml新手。我在OCaml中编写了一个简单的程序来生成公平和正方数(一个数字是回文和另一个回文的平方, 更多详情:https://code.google.com/codejam/contest/2270488/dashboard#s=p2 )如下:
更新1:优化算法(在我的笔记本电脑上大约需要20秒):
open Printf;;
let rec _is_palindrome s i j =
i >= j ||
(s.[i] = s.[j] &&
_is_palindrome s (i + 1) (j - 1))
;;
let is_palindrome s =
let sl = String.length s in
sl > 0 && (_is_palindrome s 0 (sl - 1))
;;
let rec del_zeros s =
let sl = String.length s in
if (sl < 1) then
s
else
(if s.[0] = '0' then
del_zeros (String.sub s 1 (sl - 1))
else
s
)
;;
let c2i c =
Char.code c - Char.code '0'
;;
let i2c i =
Char.chr (i + Char.code '0')
;;
(* only for finding fair and square numbers *)
let square s =
let slen = String.length s in
if slen < 1 then
""
else
let reslen = 2 * slen in
let t = ref 0 in
t := 0;
(* fast check *)
(let i = reslen/2 in
for j = (slen - 1) downto 0 do
if (i - 1 - j) >= 0 && (i - 1 - j) < slen then
t := !t + (c2i s.[j]) * (c2i s.[i - 1 - j]);
done;
if !t > 9 then
(* jump out *)
raise (Invalid_argument "carry");
);
(let res = String.make reslen '0' in
(* do the square cal now *)
for i = (reslen - 1) downto 1 do
t := 0;
for j = (slen - 1) downto 0 do
if (i - 1 - j) >= 0 && (i - 1 - j) < slen then
t := !t + (c2i s.[j]) * (c2i s.[i - 1 - j]);
done;
if !t > 9 then
(* jump out *)
raise (Invalid_argument "carry");
res.[i] <- i2c !t;
done;
del_zeros res
);
;;
let rec check_fs fsns p =
try let sq = square p in
if (is_palindrome sq) then
sq :: fsns
else
fsns
with Invalid_argument "carry" ->
fsns
;;
(* build the fair and square number list *)
(* dfs *)
let rec create_fair_square_nums fsns p sum max_num_digs =
let l = String.length p in
if l > max_num_digs || sum > 9 then
fsns
else
let fsns = create_fair_square_nums fsns ("0" ^ p ^ "0") sum max_num_digs in
let fsns = create_fair_square_nums fsns ("1" ^ p ^ "1") (sum + 1) max_num_digs in
let fsns = create_fair_square_nums fsns ("2" ^ p ^ "2") (sum + 4) max_num_digs in
let fsns = create_fair_square_nums fsns ("3" ^ p ^ "3") (sum + 9) max_num_digs in
let fsns = check_fs fsns p in
fsns
;;
let rec print_fsns fsns =
List.iter (fun s -> printf "%s " s) fsns;
printf "\n"
;;
let num_str_cmp s1 s2 =
let len1 = String.length s1 in
let len2 = String.length s2 in
match (len1 - len2) with
| 0 ->
String.compare s1 s2
| cmp -> cmp
;;
(* works *)
let max_dig = 51;;
let fsns =
let fsns = create_fair_square_nums [] "" 0 max_dig in
let fsns = create_fair_square_nums fsns "0" 0 max_dig in
let fsns = create_fair_square_nums fsns "1" 1 max_dig in
let fsns = create_fair_square_nums fsns "2" 4 max_dig in
create_fair_square_nums fsns "3" 9 max_dig
;;
let fsns = List.sort num_str_cmp fsns;;
print_fsns fsns;;
我的原始代码(天真的解决方案,太慢了):
open Printf;;
let rec _is_palindrome s i j =
if i < j then
if s.[i] = s.[j] then
_is_palindrome s (i + 1) (j - 1)
else
false
else
true
;;
let is_palindrome s =
if (String.length s < 1) then
false
else
_is_palindrome s 0 ((String.length s) - 1)
;;
let rec del_zeros s =
let sl = String.length s in
if (sl < 1) then
s
else
(if s.[0] = '0' then
del_zeros (String.sub s 1 (sl - 1))
else
s
)
;;
let c2i c =
Char.code c - Char.code '0'
;;
let i2c i =
Char.chr (i + Char.code '0')
;;
(* only positive number *)
let square s =
(* including the carry dig *)
let slen = String.length s in
let res = (
if slen > 0 then
let reslen = 2 * slen in
let res = String.make reslen '0' in
let t = ref 0 in
for i = (reslen - 1) downto 1 do
t := c2i (res.[i]);
for j = (slen - 1) downto 0 do
if (i - 1 - j) >= 0 && (i - 1 - j) < slen then
(t := !t + (c2i s.[j]) * (c2i s.[i - 1 - j]);
(* printf "%d, %d: %d\n" j (i - 1 - j) !t; *) )
done;
(* printf "%d: %d\n" i !t; *)
if !t > 9 then
(res.[i - 1] <-
Char.chr (Char.code res.[i - 1] + (!t / 10));
t := !t mod 10
);
res.[i] <- i2c !t;
done;
res;
else
""
) in
(* printf "square %s -> %s\n" s res; *)
del_zeros res
;;
let extend_palindrome new_ps n =
("0" ^ n ^ "0") ::
("1" ^ n ^ "1") ::
("2" ^ n ^ "2") ::
("3" ^ n ^ "3") ::
new_ps
;;
let rec extend_palindromes new_ps ps =
match ps with
| [] -> new_ps
| h :: t ->
let new_ps = extend_palindrome new_ps h in
extend_palindromes new_ps t
;;
let rec check_fs fsns ps =
match ps with
| [] -> fsns
| h :: t ->
let sq = square h in
if (is_palindrome sq) then
check_fs (sq :: fsns) t
else
check_fs fsns t
;;
(* build the fair and square number list *)
let rec create_fair_square_nums fsns ps max_num_digs =
match ps with
| h :: t ->
if String.length h > max_num_digs then
fsns
else
let ps = extend_palindromes [] ps in
let fsns = check_fs fsns ps in
create_fair_square_nums fsns ps max_num_digs
| [] ->
raise (Invalid_argument "fsn should not be []")
;;
let rec print_fsns fsns =
List.iter (fun s -> printf "%s " s) fsns;
printf "\n"
;;
let num_str_cmp s1 s2 =
let len1 = String.length s1 in
let len2 = String.length s2 in
match (len1 - len2) with
| 0 ->
String.compare s1 s2
| cmp -> cmp
;;
(* works *)
let max_dig = 50;;
let fsns =
let fsns0 = create_fair_square_nums [] [""] max_dig in
let fsns1 = create_fair_square_nums [] ["0"; "1"; "2"; "3"] max_dig in
(* print_fsns fsns0;
print_fsns fsns1; *)
["0"; "1"; "4"; "9"] @ fsns0 @ fsns1
;;
(* print_fsns fsns;; *)
let fsns = List.sort num_str_cmp fsns;;
print_fsns fsns;;
此代码生成10 ^ 100之内的公平和平方数字。
此代码应该有一些(或许多)有关性能的问题。在我杀了它之前它跑了30多分钟。当max_dig = 14时,它快速完成(<1分钟)。
任何关于改进此代码或批评它的建议都是受欢迎的。
答案 0 :(得分:4)
这可能是许多其他问题中的一个问题(并且可能是可以忽视的) 在您的用例中,您应该剖析以查找out),但此代码 片段已经有算法问题:
let rec del_zeros s =
let sl = String.length s in
if (sl < 1) then
s
else
(if s.[0] = '0' then
del_zeros (String.sub s 1 (sl - 1))
else
s
)
;;
String.sub在长度参数中是线性的(在内存中,
因此,时间),所以整个函数是二次的:del_zeros
(String.make 50_000 '0')
可能会很慢。
要有效地编写此代码,您应该收集保留的字符 在列表累加器中,随时累积总长度,以及 最后创建一个大小合适的字符串并写下这些字符 在它。
作为近似值,使用Buffer
的自然代码已经是
合理有效(这就是我建议平常写的东西
应用程序,但可能不在算法竞赛中,如果那是它的一部分
关键路径):
let del_zeros s =
let buf = Buffer.create (String.length s) in
for i = 0 to (String.length s) - 1 do
if s.[i] <> '0' then Buffer.add_char buf s.[i]
done;
Buffer.contents buf
答案 1 :(得分:3)
在处理大整数时,你可以使用Big_int
模块,它比你的方形自定义实现更快(并且它也可以节省你很多时间编写代码)。
此外,写if a then b else false
只是简单地写a && b
是不好的风格。