Click here for problems-> http://nabetani.sakura.ne.jp/hena/ord15subpalin/
Compare the two letters, one letter at a time from the combination of the first and last letters, If the n1st and n2nd characters are the same, increase the length (ans) by 2. Look for the same character between n1 + 1 and n2-1. (Loop in the for loop of the loop function) If the last nth and n + 1th are not the same, add 1 to the length. At first it was late without pruning in Julia, so I tried it with Nim and Ruby, After all, if the sum of the current length (ans) and the number of remaining characters is greater than the maximum length (fans) so far Only (fans <j-i + ans + 1) do a simple branch hunting to proceed with the search, If I put a break after the loop in the for loop, it became much faster. Julia
#Julia v 0.6-1.4
inp="0 I_believe_you_can_solve 9 evo_n_ove
1 a 1 a
2 aa 2 aa
3 aaa 3 aaa
4 ab 1 b
5 aabb 2 bb
6 ABBA 4 ABBA
7 Abba 2 bb
8 1234567890 1 0
9 1234567890987654321 19 1234567890987654321
10 abcdcba 7 abcdcba
11 0a1b2c3d4c5b6a7 7 abc4cba
12 abcdcba0123210 7 0123210
13 abcdcba_123210 7 abcdcba
14 _bcdcba0123210 7 0123210
15 abcddcba0123210 8 abcddcba
16 abcdcba01233210 8 01233210
17 a0bc1dc2ba3210a 9 a0123210a
18 a0bc1ddc2ba3210 8 0bcddcb0
19 3a0bc1ddc2ba3210 10 3abcddcba3
20 11oooo1111o1oo1o111ooooooooooo 20 oooo111o1oo1o111oooo
21 11o1111o1111oo11ooo11111ooo1oo 20 o1o1111oo11oo1111o1o
22 111111oo11o111ooo1o1ooo11ooo1o 21 1ooo11ooo1o1ooo11ooo1
23 11o1o1o11oo11o11oo111o1o1o11oo 27 11o1o1o11oo11o11oo11o1o1o11
24 oo111o1o11o1oo1ooo11o1o11o1o1o 27 oo111o1o11ooo1ooo11o1o111oo
25 1o1oo11111o1o1oo1o1o1111oo1o1o 28 1o1oo1111o1o1oo1o1o1111oo1o1
26 QQooooQooooQQyQoyQQQyyyyQQoyoy 15 ooQQyyQQQyyQQoo
27 oQoooQooooQyoyQoyoyyyQQyQQQQoQ 16 QoQQyQyyyyQyQQoQ
28 yyyyyooyQyyyoyyQyyooyQoQoQQoQy 17 yooQyoyyQyyoyQooy
29 yyQoyQoyyQyQQoyooooyyQQyQyooQy 24 yQooyQyQQyooooyQQyQyooQy
30 oQQooQoQyQQoyoQQoQyQyQyQoQoooo 24 oooQoQyQQyoQQoyQQyQoQooo
31 oQQyQQyyQyQQoooyQQyyyQQQyyQQoy 25 oQQyQQyyyQQoooQQyyyQQyQQo
32 WAk9iHI4jVDlStyudwTNqE138kwan2 3 wkw
33 c43fIu1Mlz0K9hEVOgGcUdbeB5ksa7 3 cGc
34 CAbYcW5VqHjzFdIkH_61PM0TsyRuie 3 HkH
35 jInQnUvSayrJTsQWujovbbqW0STvoj 10 jvTWbbWTvj
36 iZDrvpUKgtj3FrZsZ4CLjrEgUdZzQG 11 ZUgrZsZrgUZ
37 ROgYUOu6er_DA7nB6UGquO1GUHC62R 11 RUOu6B6uOUR
38 Oh_be_a_fine_girl_kiss_me 9 e_i_l_i_e
39 8086_6502_6809_Z80 11 8086_2_6808
40 xcode_visualstudio_eclipse 11 ce_iutui_ec
41 word_excel_powerpoint_outlook 9 ol_opo_lo
42 abcdefghijklmnopqrstuvwxyz0123 1 3"
function main()
local fans
function loop(str,ans)
if str !=""
for i in 1:length(str)-1
for j in length(str):-1:i+1
if str[j]==str[i] && fans<j-i+ans+1
loop(str[i+1:j-1],ans+2)
break
end
end
end
ans +=1
end
if fans<ans
fans=ans
end
end
str0=split(inp,"\n")
for s0 in str0
s1=split(s0," ")
fans=0
ans=0
loop(strip(s1[2]),ans)
if fans==parse(Int,s1[3])
sel="pass"
else
sel="fail"
end
print(sel," ",fans,"\n")
end
end
@time main()
Nim Sometimes it's not much faster than Julia, but in this problem (code) it's orders of magnitude faster.
#Nim v 0.194
import os,strutils,sequtils,times
import math
const inp="""0 I_believe_you_can_solve 9 evo_n_ove
1 a 1 a
2 aa 2 aa
3 aaa 3 aaa
4 ab 1 b
5 aabb 2 bb
6 ABBA 4 ABBA
7 Abba 2 bb
8 1234567890 1 0
9 1234567890987654321 19 1234567890987654321
10 abcdcba 7 abcdcba
11 0a1b2c3d4c5b6a7 7 abc4cba
12 abcdcba0123210 7 0123210
13 abcdcba_123210 7 abcdcba
14 _bcdcba0123210 7 0123210
15 abcddcba0123210 8 abcddcba
16 abcdcba01233210 8 01233210
17 a0bc1dc2ba3210a 9 a0123210a
18 a0bc1ddc2ba3210 8 0bcddcb0
19 3a0bc1ddc2ba3210 10 3abcddcba3
20 11oooo1111o1oo1o111ooooooooooo 20 oooo111o1oo1o111oooo
21 11o1111o1111oo11ooo11111ooo1oo 20 o1o1111oo11oo1111o1o
22 111111oo11o111ooo1o1ooo11ooo1o 21 1ooo11ooo1o1ooo11ooo1
23 11o1o1o11oo11o11oo111o1o1o11oo 27 11o1o1o11oo11o11oo11o1o1o11
24 oo111o1o11o1oo1ooo11o1o11o1o1o 27 oo111o1o11ooo1ooo11o1o111oo
25 1o1oo11111o1o1oo1o1o1111oo1o1o 28 1o1oo1111o1o1oo1o1o1111oo1o1
26 QQooooQooooQQyQoyQQQyyyyQQoyoy 15 ooQQyyQQQyyQQoo
27 oQoooQooooQyoyQoyoyyyQQyQQQQoQ 16 QoQQyQyyyyQyQQoQ
28 yyyyyooyQyyyoyyQyyooyQoQoQQoQy 17 yooQyoyyQyyoyQooy
29 yyQoyQoyyQyQQoyooooyyQQyQyooQy 24 yQooyQyQQyooooyQQyQyooQy
30 oQQooQoQyQQoyoQQoQyQyQyQoQoooo 24 oooQoQyQQyoQQoyQQyQoQooo
31 oQQyQQyyQyQQoooyQQyyyQQQyyQQoy 25 oQQyQQyyyQQoooQQyyyQQyQQo
32 WAk9iHI4jVDlStyudwTNqE138kwan2 3 wkw
33 c43fIu1Mlz0K9hEVOgGcUdbeB5ksa7 3 cGc
34 CAbYcW5VqHjzFdIkH_61PM0TsyRuie 3 HkH
35 jInQnUvSayrJTsQWujovbbqW0STvoj 10 jvTWbbWTvj
36 iZDrvpUKgtj3FrZsZ4CLjrEgUdZzQG 11 ZUgrZsZrgUZ
37 ROgYUOu6er_DA7nB6UGquO1GUHC62R 11 RUOu6B6uOUR
38 Oh_be_a_fine_girl_kiss_me 9 e_i_l_i_e
39 8086_6502_6809_Z80 11 8086_2_6808
40 xcode_visualstudio_eclipse 11 ce_iutui_ec
41 word_excel_powerpoint_outlook 9 ol_opo_lo
42 abcdefghijklmnopqrstuvwxyz0123 1 3"""
proc main()=
var fans=0
proc loop(str:string,ans:int)=
var ans1=ans
if str != "":
for i in 0..len(str)-2:
for j in countdown(len(str)-1,i+1):
if str[j]==str[i] and fans<j-i+ans1+1:
loop(str[i+1..j-1],ans1+2)
break
ans1=ans1+1
if fans<ans1:
fans=ans1
for s0 in inp.splitLines :
var s1=s0.split(" ")
fans=0
loop(s1[1].strip,0)
var sel=""
if fans==s1[2].strip.parseInt:
sel="pass"
else:
sel="fail"
echo(sel," ",fans)
var t = cpuTime()
main()
echo cpuTime()-t
Ruby I feel that there were more Ruby users among the participants in "How to write". I'm reading the first edition of "Object-Oriented Scripting Language Ruby" I had the impression that Delphi is more convenient for GUI programs and it is slower to solve puzzles. I used to use it instead of awk once in a while. However, due to another problem, it was considerably faster than Julia when I wrote it with reference to the code of other people, so I also tried this time. This problem wasn't very early before pruning, but after pruning it's a good match with Julia. From the point of view of ruby users, it may be old-fashioned and redundant writing, but I will post the pruned one.
#ruby v 2.5.5
require 'benchmark'
$inp="0 I_believe_you_can_solve 9 evo_n_ove
1 a 1 a
2 aa 2 aa
3 aaa 3 aaa
4 ab 1 b
5 aabb 2 bb
6 ABBA 4 ABBA
7 Abba 2 bb
8 1234567890 1 0
9 1234567890987654321 19 1234567890987654321
10 abcdcba 7 abcdcba
11 0a1b2c3d4c5b6a7 7 abc4cba
12 abcdcba0123210 7 0123210
13 abcdcba_123210 7 abcdcba
14 _bcdcba0123210 7 0123210
15 abcddcba0123210 8 abcddcba
16 abcdcba01233210 8 01233210
17 a0bc1dc2ba3210a 9 a0123210a
18 a0bc1ddc2ba3210 8 0bcddcb0
19 3a0bc1ddc2ba3210 10 3abcddcba3
20 11oooo1111o1oo1o111ooooooooooo 20 oooo111o1oo1o111oooo
21 11o1111o1111oo11ooo11111ooo1oo 20 o1o1111oo11oo1111o1o
22 111111oo11o111ooo1o1ooo11ooo1o 21 1ooo11ooo1o1ooo11ooo1
23 11o1o1o11oo11o11oo111o1o1o11oo 27 11o1o1o11oo11o11oo11o1o1o11
24 oo111o1o11o1oo1ooo11o1o11o1o1o 27 oo111o1o11ooo1ooo11o1o111oo
25 1o1oo11111o1o1oo1o1o1111oo1o1o 28 1o1oo1111o1o1oo1o1o1111oo1o1
26 QQooooQooooQQyQoyQQQyyyyQQoyoy 15 ooQQyyQQQyyQQoo
27 oQoooQooooQyoyQoyoyyyQQyQQQQoQ 16 QoQQyQyyyyQyQQoQ
28 yyyyyooyQyyyoyyQyyooyQoQoQQoQy 17 yooQyoyyQyyoyQooy
29 yyQoyQoyyQyQQoyooooyyQQyQyooQy 24 yQooyQyQQyooooyQQyQyooQy
30 oQQooQoQyQQoyoQQoQyQyQyQoQoooo 24 oooQoQyQQyoQQoyQQyQoQooo
31 oQQyQQyyQyQQoooyQQyyyQQQyyQQoy 25 oQQyQQyyyQQoooQQyyyQQyQQo
32 WAk9iHI4jVDlStyudwTNqE138kwan2 3 wkw
33 c43fIu1Mlz0K9hEVOgGcUdbeB5ksa7 3 cGc
34 CAbYcW5VqHjzFdIkH_61PM0TsyRuie 3 HkH
35 jInQnUvSayrJTsQWujovbbqW0STvoj 10 jvTWbbWTvj
36 iZDrvpUKgtj3FrZsZ4CLjrEgUdZzQG 11 ZUgrZsZrgUZ
37 ROgYUOu6er_DA7nB6UGquO1GUHC62R 11 RUOu6B6uOUR
38 Oh_be_a_fine_girl_kiss_me 9 e_i_l_i_e
39 8086_6502_6809_Z80 11 8086_2_6808
40 xcode_visualstudio_eclipse 11 ce_iutui_ec
41 word_excel_powerpoint_outlook 9 ol_opo_lo
42 abcdefghijklmnopqrstuvwxyz0123 1 3"
def main()
def loop(str,ans)
if str !=""
for i in 0..str.length-2
(str.length-1).downto(i+1) {
|j|
if str[j]==str[i] && $fans<j-i+ans+1
loop(str[i+1..j-1],ans+2)
break
end
}
end
ans +=1#*string(str[1])
end
if $fans<ans
$fans=ans
end
end
$inp.lines do |s0|
s1=s0.split(" ")
$fans=0
ans=0
loop(s1[1].strip,ans)
if $fans==s1[2].to_i
sel="pass"
else
sel="fail"
end
print(sel," ",$fans,"\n")
end
end
time = Benchmark.realtime do
main()
end
p time
Another solution (Prolog) This is a method to find the longest combination of the same character in the character string opposite to the original character string. Memoization recursion. If the n1 and n2 of the two strings are the same, increase the length R by 1 and compare the n1 + 1 and n2 + 1th. If they are different, when comparing the n1 + 1th and n2nd, and when comparing the n1st and n2 + 1th Add the longer one and R. If you write it like a code f (n1, n2) = if n1 == n2: f (n1 + 1, n2 + 1) + 1 # solve / 4 4th section else: max (f (n1 + 1, n2), f (n1, n2 + 1)) # solve / 4 5th section For the memoization part, use functor to make a memo of the two-dimensional array (F), In the third section of solve / 4, if it is written in the memo (nonvar (R1)), that value is used. data / 4 is used for writing and reading memos.
%SWI-Prolog v 7.4.2,7.6.4
%start.
%:-initialization(start). %ideone
twoar(_,0,_). % two dimensional array
twoar(F,X,Y):-arg(X,F,V),functor(V,y,Y),X1 is X-1,twoar(F,X1,Y).
data(F,X,Y,V):-arg(X,F,V1),arg(Y,V1,V).
solve([],_,_,0).
solve(_,[],_,0).
solve(L1,L2,F,R):-
length(L1,N1),length(L2,N2),data(F,N1,N2,R1),nonvar(R1),!,R=R1.
solve([H1|T1],[H2|T2],F,R):-
H1==H2,!,length([H1|T1],N1),length([H2|T2],N2),
solve(T1,T2,F,R1),R is R1+1,data(F,N1,N2,R).
solve([H1|T1],[H2|T2],F,R):-
length([H1|T1],N1),length([H2|T2],N2),
solve([H1|T1],T2,F,R1),solve(T1,[H2|T2],F,R2),R is max(R1,R2),data(F,N1,N2,R).
start:-str(S),split_string(S,"\s,\n","\s",L),pre(L),!.
disp(R,Q):-
number_string(R,R1),(R1==Q->Str="pass ";Str=" fail "),
write(" "),write(Str),writeln(R).
pre([]).
pre([_,B,C,_|T]):-
atom_chars(B,L),reverse(L,L1),length(L,M),M1 is M+5,functor(F,x,M1),twoar(F,M1,M1),
solve(L,L1,F,R),disp(R,C),pre(T).
str("0 I_believe_you_can_solve 9 evo_n_ove
1 a 1 a
2 aa 2 aa
3 aaa 3 aaa
4 ab 1 b
5 aabb 2 bb
6 ABBA 4 ABBA
7 Abba 2 bb
8 1234567890 1 0
9 1234567890987654321 19 1234567890987654321
10 abcdcba 7 abcdcba
11 0a1b2c3d4c5b6a7 7 abc4cba
12 abcdcba0123210 7 0123210
13 abcdcba_123210 7 abcdcba
14 _bcdcba0123210 7 0123210
15 abcddcba0123210 8 abcddcba
16 abcdcba01233210 8 01233210
17 a0bc1dc2ba3210a 9 a0123210a
18 a0bc1ddc2ba3210 8 0bcddcb0
19 3a0bc1ddc2ba3210 10 3abcddcba3
20 11oooo1111o1oo1o111ooooooooooo 20 oooo111o1oo1o111oooo
21 11o1111o1111oo11ooo11111ooo1oo 20 o1o1111oo11oo1111o1o
22 111111oo11o111ooo1o1ooo11ooo1o 21 1ooo11ooo1o1ooo11ooo1
23 11o1o1o11oo11o11oo111o1o1o11oo 27 11o1o1o11oo11o11oo11o1o1o11
24 oo111o1o11o1oo1ooo11o1o11o1o1o 27 oo111o1o11ooo1ooo11o1o111oo
25 1o1oo11111o1o1oo1o1o1111oo1o1o 28 1o1oo1111o1o1oo1o1o1111oo1o1
26 QQooooQooooQQyQoyQQQyyyyQQoyoy 15 ooQQyyQQQyyQQoo
27 oQoooQooooQyoyQoyoyyyQQyQQQQoQ 16 QoQQyQyyyyQyQQoQ
28 yyyyyooyQyyyoyyQyyooyQoQoQQoQy 17 yooQyoyyQyyoyQooy
29 yyQoyQoyyQyQQoyooooyyQQyQyooQy 24 yQooyQyQQyooooyQQyQyooQy
30 oQQooQoQyQQoyoQQoQyQyQyQoQoooo 24 oooQoQyQQyoQQoyQQyQoQooo
31 oQQyQQyyQyQQoooyQQyyyQQQyyQQoy 25 oQQyQQyyyQQoooQQyyyQQyQQo
32 WAk9iHI4jVDlStyudwTNqE138kwan2 3 wkw
33 c43fIu1Mlz0K9hEVOgGcUdbeB5ksa7 3 cGc
34 CAbYcW5VqHjzFdIkH_61PM0TsyRuie 3 HkH
35 jInQnUvSayrJTsQWujovbbqW0STvoj 10 jvTWbbWTvj
36 iZDrvpUKgtj3FrZsZ4CLjrEgUdZzQG 11 ZUgrZsZrgUZ
37 ROgYUOu6er_DA7nB6UGquO1GUHC62R 11 RUOu6B6uOUR
38 Oh_be_a_fine_girl_kiss_me 9 e_i_l_i_e
39 8086_6502_6809_Z80 11 8086_2_6808
40 xcode_visualstudio_eclipse 11 ce_iutui_ec
41 word_excel_powerpoint_outlook 9 ol_opo_lo
42 abcdefghijklmnopqrstuvwxyz0123 1 3").
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