Tulis program yang menampilkan jumlah total karakter dan frekuensi setiap karakter dalam sumbernya dan hasilnya. Anda harus mengikuti format yang diilustrasikan dalam contoh ini.
Contoh
Jika kode Anda
abb1
Hasilnya harus
My source has 4 characters.
1 is "a"
2 are "b"
1 is "1"
Besides unquoted numbers, my output has 383 characters.
34 are "
"
79 are " "
63 are """
2 are "'"
2 are ","
4 are "."
2 are "1"
2 are "B"
2 are "I"
2 are "M"
39 are "a"
4 are "b"
6 are "c"
4 are "d"
38 are "e"
3 are "g"
5 are "h"
4 are "i"
4 are "m"
3 are "n"
8 are "o"
3 are "p"
2 are "q"
38 are "r"
12 are "s"
8 are "t"
7 are "u"
3 are "y"
It's good to be a program.
(Output harus menuju stdout.)
Perhatikan, misalnya, bahwa output mengandung dua huruf kapital m. Satu untuk Mydan satu untuk 2 are "M". Ini harus berlaku untuk semua karakter sehingga output tidak bertentangan dengan cara apa pun.
Angka yang tidak dikutip diabaikan dalam output untuk menghindari set frekuensi yang tidak memuaskan. Misalnya, 1 is "1"salah jika keduanya dihitung. Seharusnya membaca 2 are "1", tetapi kemudian hanya ada satu 1 lagi.
Klarifikasi Format
"is" harus digunakan untuk kemunculan karakter tunggal.
"are" harus digunakan untuk kemunculan beberapa karakter.
"is" seharusnya tidak pernah muncul dalam daftar karakter keluaran karena itu akan berlebihan.
1 is 'Z'mengacu pada Z itu sendiri, sehingga seluruh baris dapat dihapus.Tiga frasa kalimat lengkap harus muncul sesuai dengan daftar frekuensi karakter di antaranya (seperti yang ditunjukkan contoh). Jadi output Anda akan mulai dengan
My source...dan diakhiri dengan...be a program.. Perhatikan bahwa tidak ada baris baru di akhir output.Daftar frekuensi karakter itu sendiri mungkin berurutan.
Baris baru dihitung sebagai satu karakter (seandainya itu adalah \ r \ n).
Pemeriksa format
Skrip Python berikut mengambil kode Anda dan hasilnya sebagai string dan menyatakan bahwa output tidak memiliki kontradiksi. Ini memberikan pesan kesalahan yang berguna jika ada sesuatu yang salah. Anda dapat menjalankannya secara online di http://ideone.com/6H0ldu dengan memalsunya, mengganti string KODE dan OUTPUT, kemudian menjalankannya. Itu tidak akan pernah memberikan positif atau negatif palsu (dengan asumsi kesalahan bebas).
#Change the CODE and OUTPUT strings to test your program
CODE = r'''abb1'''
OUTPUT = r'''My source has 4 characters.
1 is "a"
2 are "b"
1 is "1"
Besides unquoted numbers, my output has 383 characters.
34 are "
"
79 are " "
63 are """
2 are "'"
2 are ","
4 are "."
2 are "1"
2 are "B"
2 are "I"
2 are "M"
39 are "a"
4 are "b"
6 are "c"
4 are "d"
38 are "e"
3 are "g"
5 are "h"
4 are "i"
4 are "m"
3 are "n"
8 are "o"
3 are "p"
2 are "q"
38 are "r"
12 are "s"
8 are "t"
7 are "u"
3 are "y"
It's good to be a program.'''
#######################################################
import re
amountPattern = r'(\d+) (is|are) "(.)"\n'
class IntrospectionException(Exception):
pass
def getClaimedAmounts(string, errorOnIs):
groups = re.findall(amountPattern, string, re.DOTALL)
for amount, verb, char in groups:
if verb == 'is':
if errorOnIs:
raise IntrospectionException('\'1 is "%s"\' is unnecessary' % char)
elif amount != '1':
raise IntrospectionException('At "%s", %s must use "are"' % (char, amount))
elif verb == 'are' and amount == '1':
raise IntrospectionException('At "%s", 1 must use "is"' % char)
amounts = {}
for amount, verb, char in groups:
if char in amounts:
raise IntrospectionException('Duplicate "%s" found' % char)
amounts[char] = int(amount)
return amounts
def getActualAmounts(string):
amounts = {}
for char in string:
if char in amounts:
amounts[char] += 1
else:
amounts[char] = 1
return amounts
def compareAmounts(claimed, actual):
for char in actual:
if char not in claimed:
raise IntrospectionException('The amounts list is missing "%s"' % char)
for char in actual: #loop separately so missing character errors are all found first
if claimed[char] != actual[char]:
raise IntrospectionException('The amount of "%s" characters is %d, not %d' % (char, actual[char], claimed[char]))
if claimed != actual:
raise IntrospectionException('The amounts are somehow incorrect')
def isCorrect(code, output):
p1 = r'^My source has (\d+) characters\.\n'
p2 = r'Besides unquoted numbers, my output has (\d+) characters\.\n'
p3 = r"It's good to be a program\.$"
p4 = '%s(%s)*%s(%s)*%s' % (p1, amountPattern, p2, amountPattern, p3)
for p in [p1, p2, p3, p4]:
if re.search(p, output, re.DOTALL) == None:
raise IntrospectionException('Did not match the regex "%s"' % p)
claimedCodeSize = int(re.search(p1, output).groups()[0])
actualCodeSize = len(code)
if claimedCodeSize != actualCodeSize:
raise IntrospectionException('The code length is %d, not %d' % (actualCodeSize, claimedCodeSize))
filteredOutput = re.sub(r'([^"])\d+([^"])', r'\1\2', output)
claimedOutputSize = int(re.search(p2, output).groups()[0])
actualOutputSize = len(filteredOutput)
if claimedOutputSize != actualOutputSize:
raise IntrospectionException('The output length (excluding unquoted numbers) is %d, not %d' % (actualOutputSize, claimedOutputSize))
splitIndex = re.search(p2, output).start()
claimedCodeAmounts = getClaimedAmounts(output[:splitIndex], False)
actualCodeAmounts = getActualAmounts(code)
compareAmounts(claimedCodeAmounts, actualCodeAmounts)
claimedOutputAmounts = getClaimedAmounts(output[splitIndex:], True)
actualOutputAmounts = getActualAmounts(filteredOutput)
compareAmounts(claimedOutputAmounts, actualOutputAmounts)
def checkCorrectness():
try:
isCorrect(CODE, OUTPUT)
print 'Everything is correct!'
except IntrospectionException as e:
print 'Failed: %s.' % e
checkCorrectness()
Mencetak gol
Ini adalah kode-golf. Kiriman dengan karakter paling sedikit menang. Pengajuan harus melewati pemeriksa format agar valid. Celah standar berlaku, meskipun Anda dapat membaca kode sumber Anda sendiri dan / atau membuat hardcode keluaran Anda .
r'''CODE''').