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Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
gl = ["a", "i", "e", "o", "u"]
t = []
i = 0
while i + 2 < len(s):
good = 0
if s[i] in gl or s[i + 1] in gl or s[i + 2] in gl or s[i] == s[i + 1] == s[i + 2]:
good = 1
t.append(s[i])
if not good:
t.append(s[i + 1])
t.append(" ")
i += 1
i += 1
while i < len(s):
t.append(s[i])
i += 1
print("".join(t)) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER VAR NUMBER WHILE VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | def glas(gl, s):
if s[0] in gl or s[1] in gl or s[2] in gl:
return 1
else:
return 0
def right(gl, s):
if s[0] == s[1] and s[0] == s[2] or glas(gl, s):
return 1
else:
return 0
s = input()
i = len(s)
gl = "aeoiu"
k = 0
while k <= i - 3:
s1 = s[k : k + 3]
if not right(gl, s1):
s = s[: k + 2] + " " + s[k + 2 :]
i += 1
k += 3
else:
k += 1
print(s) | FUNC_DEF IF VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER FUNC_CALL VAR VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
s1 = ""
i = 0
while i < len(s) - 2:
if (
(s[i] != s[i + 1] or s[i + 2] != s[i + 1] or s[i] != s[i + 2])
and s[i] != "i"
and s[i] != "a"
and s[i] != "e"
and s[i] != "o"
and s[i] != "u"
and s[i + 1] != "i"
and s[i + 1] != "a"
and s[i + 1] != "e"
and s[i + 1] != "o"
and s[i + 1] != "u"
and s[i + 2] != "i"
and s[i + 2] != "a"
and s[i + 2] != "e"
and s[i + 2] != "o"
and s[i + 2] != "u"
):
s1 = s[0 : i + 2] + " " + s[i + 2 : len(s)]
s = s1
i += 3
else:
s1 = s
i += 1
if s1 == "":
s1 = s
print(s1) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR STRING VAR VAR STRING VAR VAR STRING VAR VAR STRING VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR STRING ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | def puzzle():
st = input()
a = ""
b = ""
c = ""
res = ""
vowels = "aiueo "
for i in range(len(st)):
res += c
c = b
b = a
a = st[i]
if a not in vowels and b not in vowels and c not in vowels:
if not a == b == c:
res += c
c = b
b = " "
res += c + b + a
print(res)
puzzle() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR VAR VAR VAR IF VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR STRING VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
n = len(s)
if n == 1 or n == 2:
print(s)
else:
a = ["a", "e", "i", "o", "u"]
q, w = "", ""
if s[0] not in a:
q = s[0]
if s[1] not in a:
w = s[1]
res = s[:2]
for i in range(2, n):
e = s[i]
if e not in a:
if q != "" and w != "":
if q == w == e:
q = w
res += e
else:
res += " " + e
q, w = "", e
else:
q, w = w, e
res += e
else:
res += e
q, w = "", ""
print(res) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR VAR STRING STRING IF VAR NUMBER VAR ASSIGN VAR VAR NUMBER IF VAR NUMBER VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR IF VAR VAR IF VAR STRING VAR STRING IF VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR BIN_OP STRING VAR ASSIGN VAR VAR STRING VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR STRING STRING EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | ans = ["a", "e", "i", "o", "u"]
s = input()
point = 0
for i in range(2, len(s)):
r = s[i] == s[i - 1] == s[i - 2]
if (
s[i] not in ans
and s[i - 1] not in ans
and s[i - 2] not in ans
and r == False
and i - 2 >= point
):
print(s[point:i], end=" ")
point = i
print(s[point:]) | ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR STRING ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | class Solver:
def main(self):
s = input()
ls = len(s)
vowels = "aeiou"
cnt = 0
breaks = []
same = 0
for i in range(0, ls):
if s[i] in vowels:
cnt = 0
same = 0
elif i > 0 and s[i - 1] == s[i]:
same += 1
cnt = cnt
else:
cnt += 1
if same + cnt >= 3 and cnt >= 2:
breaks.append(i)
cnt = 1
same = 0
lbreaks = len(breaks)
current = 0
for i in range(ls):
if current < lbreaks and i == breaks[current]:
print(" ", end="")
current += 1
print(s[i], end="")
print()
Solver().main() | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING STRING VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL FUNC_CALL VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input().strip()
gl = {"a", "i", "o", "u", "e"}
answer = ""
word = []
i = 1
while i < len(s) - 1:
if (s[i] not in gl and s[i - 1] not in gl and s[i + 1] not in gl) and len(
{s[i], s[i - 1], s[i + 1]}
) > 1:
s = s[: i + 1] + " " + s[i + 1 :]
i += 3
else:
i += 1
print(s) | ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR STRING STRING STRING STRING STRING ASSIGN VAR STRING ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | vowels = ["a", "e", "i", "o", "u"]
word = [str(i) for i in input()]
consonant_counter = 0
increment = -1
vowelchain = []
for letter in word:
increment += 1
if letter not in vowels:
vowelchain.append(letter)
consonant_counter += 1
if consonant_counter == 3 and vowelchain.count(letter) != 3:
word.insert(increment, " ")
consonant_counter = 0
vowelchain = []
elif (
consonant_counter > 3
and len(vowelchain) > 3
and vowelchain.count(letter) < 3
):
word.insert(increment, " ")
consonant_counter = 0
vowelchain = []
else:
consonant_counter = 0
vowelchain = []
print("".join(word)) | ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR LIST IF VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | import sys
vowels = set(["a", "e", "i", "o", "u"])
word = sys.stdin.readline().rstrip("\n")
correct = []
curr = []
cons = []
for i in range(len(word)):
if word[i] in vowels:
curr.append(word[i])
cons = []
else:
cons.append(word[i])
if len(set(cons)) != 1 and len(cons) > 2:
curr = "".join(curr)
correct.append(curr)
curr = []
cons = [cons[-1]]
curr.append(word[i])
if len(curr) > 0:
correct.append("".join(curr))
sys.stdout.write(" ".join(correct) + "\n") | IMPORT ASSIGN VAR FUNC_CALL VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL STRING VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | input_str = input("")
input_str_len = len(input_str)
if input_str_len > 3000 or input_str_len < 1:
print("-")
else:
vowels = {"a", "e", "i", "o", "u", " "}
last_char = ""
final_str = ""
consecutive_consonants = 0
consecutive_same_consonants_in_a_row = 0
for v in input_str:
final_str += last_char
if v in vowels:
consecutive_consonants = 0
consecutive_same_consonants_in_a_row = 0
else:
if consecutive_consonants >= 2:
if consecutive_same_consonants_in_a_row == 1 or last_char != v:
final_str += " "
consecutive_consonants = 0
consecutive_same_consonants_in_a_row = 0
if last_char == v:
consecutive_same_consonants_in_a_row += 1
else:
consecutive_same_consonants_in_a_row = 1
consecutive_consonants += 1
last_char = v
final_str += last_char
print(final_str) | ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR STRING STRING STRING STRING STRING STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF VAR NUMBER VAR VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | good = ["a", "e", "i", "o", "u"]
s = input().strip()
ans = []
nw = ""
i = 0
while i < len(s) - 2:
nw += s[i]
if not s[i] == s[i + 1] == s[i + 2]:
if not (s[i] in good or s[i + 1] in good or s[i + 2] in good):
nw += s[i + 1]
ans.append(nw)
i += 1
nw = ""
i += 1
ans.append(nw + s[i:])
print(" ".join(ans)) | ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR STRING VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
vow = ["a", "e", "i", "o", "u"]
c = 0
n = len(s)
arr = []
for i in range(0, n):
if s[i] in vow:
c = 0
continue
if s[i] not in vow:
c += 1
if c == 3:
if s[i] == s[i - 1] and s[i - 1] == s[i - 2]:
c = 2
continue
arr.append(i)
c = 1
k = 0
for i in range(len(arr)):
print(s[k : arr[i]], end=" ")
k = arr[i]
print(s[k:]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR NUMBER IF VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR STRING ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | word = list(input())
ns = 0
onesymbol = ""
onesymbolcount = 0
for i in range(len(word)):
if onesymbol == word[i] and ns == 0:
onesymbolcount += 1
continue
if word[i] in ["a", "e", "i", "o", "u"]:
ns = 0
onesymbolcount = 0
onesymbol = ""
else:
if onesymbol != "":
ns += onesymbolcount
onesymbol = word[i]
onesymbolcount = 1
if ns >= 2:
word[i - 1] += " "
ns = 0
print("".join(word)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING IF VAR STRING VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | k = 0
t = y = z = ""
for x in input():
k = 0 if x in "aeiou" else k + 1
if k > 2 and not x == y == z:
t += " "
k = 1
y, z = x, y
t += x
print(t) | ASSIGN VAR NUMBER ASSIGN VAR VAR VAR STRING FOR VAR FUNC_CALL VAR ASSIGN VAR VAR STRING NUMBER BIN_OP VAR NUMBER IF VAR NUMBER VAR VAR VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
s2 = ""
gl = "aeiou"
i = 0
while i < len(s):
s2 += s[i]
if i + 2 < len(s):
if not (s[i] in gl or s[i + 1] in gl or s[i + 2] in gl):
if not (s[i] == s[i + 1] and s[i + 1] == s[i + 2]):
s2 += s[i + 1] + " "
i += 1
i += 1
print(s2) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | import sys
MOD, mod = 1000000007, 998244353
def get_array():
return list(map(int, sys.stdin.readline().strip().split()))
def get_ints():
return map(int, sys.stdin.readline().strip().split())
def input():
return sys.stdin.readline().strip()
string = input()
n = len(string)
i = 0
ans = ""
vowel = ["a", "e", "i", "o", "u"]
while i < n - 2:
if string[i] in vowel:
ans += string[i]
i += 1
elif string[i] == string[i + 1] == string[i + 2]:
ans += string[i]
i += 1
elif string[i + 1] not in vowel and string[i + 2] not in vowel:
ans += string[i] + string[i + 1] + " "
i += 2
else:
ans += string[i]
i += 1
ans += string[i:]
print(ans) | IMPORT ASSIGN VAR VAR NUMBER NUMBER FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR LIST STRING STRING STRING STRING STRING WHILE VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR VAR VAR VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
ans = s
g = "aeiou "
i = 0
while i < len(ans) - 2:
if ans[i] == ans[i + 1] and ans[i] == ans[i + 2]:
i += 1
continue
if ans[i] not in g and ans[i + 1] not in g and ans[i + 2] not in g:
ans = ans[: i + 2] + " " + ans[i + 2 :]
i += 1
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | n = input()
n += " "
def alph(a):
flag = bool(True)
for i in ["a", "e", "i", "o", "u", " "]:
if i == a:
flag = False
if flag == False:
return 0
else:
return 1
i = 0
while i != len(n) - 2:
if alph(n[i]) == 1 and alph(n[i + 1]) == 1 and alph(n[i + 2]) == 1:
if n[i] != n[i + 1] or n[i + 1] != n[i + 2] or n[i] != n[i + 2]:
n = n[: i + 2] + " " + n[i + 2 :]
i += 1
print(n[:-2]) | ASSIGN VAR FUNC_CALL VAR VAR STRING FUNC_DEF ASSIGN VAR FUNC_CALL VAR NUMBER FOR VAR LIST STRING STRING STRING STRING STRING STRING IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
n = len(s)
alf = ["a", "e", "i", "u", "o"]
i = 0
ch = 0
while i < n:
if not s[i] in alf:
ch += 1
else:
ch = 0
if ch == 3:
if not s[i - 2] == s[i - 1] == s[i]:
print(" ", end="")
ch = 1
else:
ch = 2
print(s[i], end="")
i += 1 | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING VAR NUMBER |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
res = []
l = 0
for r in range(2, len(s)):
if (
r - l >= 2
and not any(c in "aeiou" for c in s[r - 2 : r + 1])
and s[r - 2 : r + 1].count(s[r]) < 3
):
res.append(s[l:r])
l = r
res.append(s[l:])
print(" ".join(res)) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR STRING VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | def f(x):
return x not in "aeiou"
s, res = input(), ""
i = 0
while i < len(s) - 2:
if (
f(s[i])
and f(s[i + 1])
and f(s[i + 2])
and (s[i] != s[i + 1] or s[i + 1] != s[i + 2])
):
res += s[i] + s[i + 1] + " "
i += 2
else:
res += s[i]
i += 1
res += s[i:]
print(res) | FUNC_DEF RETURN VAR STRING ASSIGN VAR VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR VAR VAR VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
k = 0
lis = ["a", "e", "i", "o", "u", " "]
x = ""
z = ""
v = ""
for j in s:
if j not in lis:
if z == j == v:
k = 2
else:
k += 1
v = z
z = j
if k == 3:
x += " "
k = 1
x += j
else:
v = z
z = j
k = 0
x += j
print(x) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST STRING STRING STRING STRING STRING STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING FOR VAR VAR IF VAR VAR IF VAR VAR VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER VAR STRING ASSIGN VAR NUMBER VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
words = ""
g = "aeiou"
i = 0
while len(s) > 2:
if s[i] not in g and s[i + 1] not in g and s[i + 2] not in g:
if s[i] == s[i + 1] == s[i + 2]:
i += 1
if i == len(s) - 2:
break
continue
words += s[: i + 2] + " "
s = s[i + 2 :]
i = 0
else:
i += 1
if i == len(s) - 2:
break
print(words + s) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
c = "aeiou"
ans = "@"
cn = 0
sm = 1
for i in s:
if i in c:
cn = 0
sm = 1
elif i == ans[-1]:
sm += 1
cn += 1
elif i != ans[-1]:
cn += 1
if cn == sm or cn < 3:
ans += i
elif cn > 2:
ans += " "
ans += i
cn = 1
sm = 1
print(ans[1:]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR IF VAR NUMBER VAR STRING VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | def check(st=str()):
tf = 0
for i in range(2):
if st[i] != st[i + 1]:
tf = 1
if tf == 1:
return True
else:
return False
st = input()
anst = ""
vowels = "a", "e", "i", "o", "u"
count = 0
i = 0
while i < len(st):
if st[i] not in vowels:
count += 1
else:
count = 0
if count == 3 and check(st[i - 2 : i + 1]):
anst += " "
count = 0
else:
anst += st[i]
if count == 3:
count = 2
i += 1
print(anst) | FUNC_DEF FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR STRING ASSIGN VAR NUMBER VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
n = len(s)
sogl = ["a", "e", "i", "o", "u"]
i = 0
ans = []
while i < n - 2:
if (
not s[i] in sogl
and not s[i + 1] in sogl
and not s[i + 2] in sogl
and not s[i] == s[i + 1] == s[i + 2]
):
ans.append(i + 2)
i += 1
i += 1
j = 0
for i in ans:
print(s[j:i], end=" ")
j = i
print(s[j:]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR STRING ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
k = 0
s1 = ""
s2 = ["", "", ""]
for a in s:
if a == "a" or a == "e" or a == "i" or a == "o" or a == "u":
s1 += a
k = 0
elif k + 1 == 3:
s2[2] = a
if s2[0] != s2[1] or s2[0] != s2[2]:
s1 = s1 + " " + a
k = 1
s2[0] = a
else:
k = 2
s1 += a
else:
s2[k] = a
k += 1
s1 += a
print(s1) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR LIST STRING STRING STRING FOR VAR VAR IF VAR STRING VAR STRING VAR STRING VAR STRING VAR STRING VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR STRING VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | vowels = ["a", "e", "i", "o", "u"]
s = input()
i = 0
l = list(s)
while i < len(l) - 2:
if l[i] not in vowels and l[i + 1] not in vowels and l[i + 2] not in vowels:
if l[i] * 3 != l[i] + l[i + 1] + l[i + 2]:
l.insert(i + 2, " ")
i += 2
i += 1
print("".join(l)) | ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR IF BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | def main():
s = input()
a = []
g = [elem for elem in "aeiou"]
p = []
for i in range(len(s)):
if s[i] not in g:
a.append(s[i])
else:
a = []
if len(a) == 3:
if a[0] != a[1] or a[1] != a[2]:
a = [a[2]]
p.append(i)
else:
a = [a[1], a[2]]
for i in range(len(s)):
if i in p:
print(" ", end="")
print(s[i], end="")
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR VAR VAR STRING ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR LIST VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING STRING EXPR FUNC_CALL VAR VAR VAR STRING EXPR FUNC_CALL VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | word = input()
l = ["a", "e", "i", "o", "u"]
curr_s = []
i = 0
for j in range(len(word)):
if word[j] not in l:
if curr_s.count(word[j]) > 1:
continue
curr_s.append(word[j])
if len(curr_s) > 2:
print(word[i:j], end=" ")
curr_s = [word[j]]
i = j
else:
curr_s = []
print(word[i:]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR STRING ASSIGN VAR LIST VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | i = input()
alphabet = [
"B",
"C",
"D",
"F",
"G",
"H",
"J",
"K",
"L",
"M",
"N",
"P",
"Q",
"R",
"S",
"T",
"V",
"W",
"X",
"Y",
"Z",
]
res = ""
flag = 0
for j in range(len(i)):
if flag > 0:
flag -= 1
res += i[j]
continue
if j != 0 and j != len(i) - 1:
if (
i[j].upper() in alphabet
and i[j - 1].upper() in alphabet
and i[j + 1].upper() in alphabet
) and not (i[j - 1] == i[j] and i[j] == i[j + 1]):
res += i[j] + " "
flag = 1
continue
else:
res += i[j]
else:
res += i[j]
print(res) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER VAR VAR VAR IF VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR VAR STRING ASSIGN VAR NUMBER VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | result, seq, same, prev = [], 0, True, ""
for i in input():
seq = 0 if i in "aeiou" else seq + 1
same = seq == 1 or same and i == prev
prev = i
if seq >= 3 and not same:
result.append(" ")
seq, same = 1, True
result.append(i)
print("".join(result)) | ASSIGN VAR VAR VAR VAR LIST NUMBER NUMBER STRING FOR VAR FUNC_CALL VAR ASSIGN VAR VAR STRING NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER VAR VAR VAR ASSIGN VAR VAR IF VAR NUMBER VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
ans = ""
vowels = {"a", "e", "i", "o", "u"}
i = 0
while i < len(s) - 2:
if s[i] in vowels:
ans += s[i]
i += 1
elif s[i] == s[i + 1] == s[i + 2]:
ans += s[i]
i += 1
elif s[i + 1] not in vowels and s[i + 2] not in vowels:
ans += s[i] + s[i + 1] + " "
i += 2
else:
ans += s[i]
i += 1
ans += s[i:]
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING STRING STRING STRING STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR VAR VAR VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = "qwrtpsdfghjklzxcvbnmy"
t = input()
i = 1
while i < len(t) - 1:
k = 0
if s.find(t[i - 1]) != -1:
k += 1
if s.find(t[i]) != -1:
k += 1
if s.find(t[i + 1]) != -1:
k += 1
if t[i - 1] == t[i] == t[i + 1]:
k = 0
if k == 3:
t = t[0 : i + 1] + " " + t[i + 1 : len(t)]
i += 1
print(t) | ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER IF FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
l = ["a", "e", "i", "o", "u"]
sp = []
for i in range(len(s)):
sp.append(s[i])
k = 0
for i in range(len(s)):
if s[i] not in l:
k += 1
if i > 1 and s[i] == s[i - 1] == s[i - 2]:
k = 2
else:
k = 0
if k == 3:
sp[i - 1] = sp[i - 1] + " "
k = 1
for i in sp:
print(i, end="") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER STRING ASSIGN VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | import sys
s = input()
p = [0] * 30000
i = 0
while i < len(s) - 2:
if (
s[i] not in "aeiou"
and s[i + 1] not in "aeiou"
and s[i + 2] not in "aeiou"
and (s[i] != s[i + 1] or s[i] != s[i + 2])
):
p[i + 1] = 1
i += 2
else:
i += 1
ans = ""
for i in range(len(s)):
ans += s[i]
if p[i]:
ans += " "
print(ans) | IMPORT ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR VAR STRING EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | a = input()
c = []
ind = []
for i in range(len(a)):
if a[i] not in "aeiou":
if len(c) > 1 and not (a[i] == c[-1] and a[i] == c[-2]):
ind.append(i)
c = []
c.append(a[i])
else:
c = []
ind.append(len(a))
print(a[: ind[0]], end=" ")
for i in range(1, len(ind)):
print(a[ind[i - 1] : ind[i]], end=" ") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER STRING FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
c = []
ans = ""
for i in s:
if i in "eioua":
c = []
else:
c.append(i)
if len(c) > 2 and len(set(c)) > 1:
ans += " "
c = [i]
ans += i
print(ans) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR STRING FOR VAR VAR IF VAR STRING ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR STRING ASSIGN VAR LIST VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | a = [
"b",
"c",
"d",
"f",
"g",
"h",
"j",
"k",
"l",
"m",
"n",
"p",
"q",
"r",
"s",
"t",
"v",
"w",
"x",
"y",
"z",
]
b = input()
n = 0
g = 0
s = []
check = False
print(b[0], end="")
for i in range(1, len(b) - 1):
if check:
print(b[i], end="")
check = False
continue
m = [b[i - 1], b[i], b[i + 1]]
if b[i - 1] in a and b[i] in a and b[i + 1] in a and len(list(set(m))) > 1:
print(b[i], end=" ")
check = True
else:
print(b[i], end="")
if len(b) > 1:
print(b[-1]) | ASSIGN VAR LIST STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER STRING FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR LIST VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | def sogl(a):
glas = ["u", "a", "i", "e", "o"]
for letter in glas:
if letter == a:
return False
return True
def xor(a, b, c):
j = 0
if a == b:
j += 1
if a == c:
j += 1
if j == 2:
return False
else:
return True
line = input()
let = 0
for i in range(0, len(line)):
if sogl(line[i]) and let < 2:
let += 1
print(line[i], end="")
elif sogl(line[i]) and let == 2 and xor(line[i], line[i - 1], line[i - 2]):
let = 1
print(" ", end="")
print(line[i], end="")
elif sogl(line[i]) and let == 2 and not xor(line[i], line[i - 1], line[i - 2]):
print(line[i], end="")
else:
let = 0
print(line[i], end="") | FUNC_DEF ASSIGN VAR LIST STRING STRING STRING STRING STRING FOR VAR VAR IF VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER IF VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING IF FUNC_CALL VAR VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR STRING STRING EXPR FUNC_CALL VAR VAR VAR STRING IF FUNC_CALL VAR VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
n = len(s)
gl = "aeiou"
i = 2
pr = lambda i: print(s[i], end="")
pr(0)
if 1 < n:
pr(1)
while i < n:
flag = 1
for j in range(i - 2, i + 1):
if s[j] in gl:
flag = 0
if len(set(s[i - 2 : i + 1])) > 1 and flag:
print(end=" ")
pr(i)
i += 1
if i < n:
pr(i)
i += 1 | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR STRING EXPR FUNC_CALL VAR NUMBER IF NUMBER VAR EXPR FUNC_CALL VAR NUMBER WHILE VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s, i = input(), 0
ans = 0
while i < len(s):
cur = []
while i < len(s) and s[i] in ["a", "e", "i", "o", "u"]:
print(s[i], end="")
i += 1
while i < len(s) and not s[i] in ["a", "e", "i", "o", "u"]:
cur += s[i]
i += 1
if len(cur) == 0:
break
j, sz = 1, 1
print(cur[0], end="")
while j < len(cur):
if sz == 0:
sz = 1
print(cur[j], end="")
elif cur[j] == cur[j - 1]:
sz += 1
print(cur[j], end="")
elif sz == 1:
print(cur[j], end="")
if j != len(cur) - 1:
print(" ", end="")
is_one = True
sz = 0
else:
print(" ", end="")
print(cur[j], end="")
is_one = True
sz = 1
j += 1 | ASSIGN VAR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR LIST WHILE VAR FUNC_CALL VAR VAR VAR VAR LIST STRING STRING STRING STRING STRING EXPR FUNC_CALL VAR VAR VAR STRING VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR LIST STRING STRING STRING STRING STRING VAR VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER STRING WHILE VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING IF VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR STRING STRING EXPR FUNC_CALL VAR VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | checking = input()
vowels = ["a", "e", "i", "o", "u"]
streak = []
new = ""
for i, letter in enumerate(checking):
if letter not in vowels:
streak.append(letter)
else:
streak = []
if len(streak) == 3:
if streak.count(letter) == 3:
streak = [letter, letter]
else:
streak = [letter]
new += " "
new += letter
print(new) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR LIST ASSIGN VAR STRING FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST VAR VAR ASSIGN VAR LIST VAR VAR STRING VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | a = input()
a += "a"
g = {"a", "e", "i", "o", "u"}
b = ""
def seach(b):
b += "a"
count = 0
i = -1
if len(set(b)) != 1:
while True:
i += 1
count += 1
if b[i] != b[i - 1]:
if count > 2:
i1 = i - count
i2 = i
if not (i1 == 0 and i2 + 1 == len(b)):
if i1 == 0:
b = b[:i2] + "|" + b[i2:]
elif i2 + 1 == len(b):
b = b[:i1] + "|" + b[i1:]
else:
b = b[:i1] + "|" + b[i1:i2] + "|" + b[i2:]
count = 0
if b[i] == "a":
break
b = b[: len(b) - 1]
b = b.replace("||", "|")
b = b + "|"
i1 = 0
i2 = 0
c = b
for i in range(len(b)):
if b[i] == "|":
n = 0
if len(set(b[i1:i])) != 1 and i - i1 > 1:
l = 2
for z in range(2, len(b[i1:i]), 2):
c = c[: i2 + l] + "|" + c[i2 + l :]
l += 3
n += 1
i2 += i + 1 - i1 + n
i1 = i + 1
c = c[: len(c) - 1]
return c
ans = ""
for i in range(len(a)):
if a[i] in g:
if len(b) > 2:
ans += seach(b)
else:
ans += b
b = ""
ans += a[i]
else:
b += a[i]
ans = ans[: len(ans) - 1]
ans = ans.replace("||", " ")
ans = ans.replace("|", " ")
print(ans) | ASSIGN VAR FUNC_CALL VAR VAR STRING ASSIGN VAR STRING STRING STRING STRING STRING ASSIGN VAR STRING FUNC_DEF VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER WHILE NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR STRING VAR VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR STRING VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR STRING VAR VAR VAR STRING VAR VAR ASSIGN VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING STRING ASSIGN VAR BIN_OP VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR STRING VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR STRING VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING STRING ASSIGN VAR FUNC_CALL VAR STRING STRING EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
c = s + ""
a = []
b = [
"b",
"c",
"d",
"f",
"g",
"h",
"j",
"k",
"l",
"m",
"n",
"p",
"q",
"r",
"s",
"t",
"v",
"w",
"x",
"y",
"z",
]
for i in range(2, len(s)):
if (
not c[i - 2] == c[i - 1] == c[i]
and c[i - 2] in b
and c[i - 1] in b
and c[i] in b
):
a.append(i)
c = c[: i - 1] + "a" + c[i:]
for i in range(len(s)):
if i in a:
print(" " + s[i], end="")
else:
print(s[i], end="") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR STRING ASSIGN VAR LIST ASSIGN VAR LIST STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER STRING VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP STRING VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
i = 0
ans = []
v = "aeiou"
dif = set()
n = 0
while i < len(s):
if s[i] not in v:
dif.add(s[i])
n += 1
elif s[i] in v:
dif = set()
n = 0
if len(dif) > 1 and n > 2:
ans.append(" ")
dif = set([s[i]])
n = 1
ans.append(s[i])
i += 1
print(*ans, sep="") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR LIST VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
if len(s) <= 2 or len(set(s)) == 1:
print(s)
exit()
d = {}
for i in range(97, 123):
x = chr(i)
if x in ["a", "e", "i", "o", "u"]:
d.update({x: False})
else:
d.update({x: True})
l, i = [], 0
while True:
if i >= len(s) - 2:
break
if d[s[i]] == True and d[s[i + 1]] == True and d[s[i + 2]] == True:
y = s[i] + s[i + 1] + s[i + 2]
if len(set(y)) > 1:
x = s[: i + 2]
l.append(x)
l.append(" ")
s = s[i + 2 :]
i = 0
else:
i = i + 1
else:
i += 1
if len(s) > 0:
l.append(s)
for i in l:
print(i, end="")
print() | ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR DICT FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR LIST STRING STRING STRING STRING STRING EXPR FUNC_CALL VAR DICT VAR NUMBER EXPR FUNC_CALL VAR DICT VAR NUMBER ASSIGN VAR VAR LIST NUMBER WHILE NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | vowels = "aeiou"
word = input()
consonants = word[0] not in vowels
duplicates = 0
result = ""
term = word[0]
for p, w in zip(word, word[1:]):
if w not in vowels:
consonants += 1
else:
consonants = duplicates = 0
if consonants > 0:
duplicates += p == w
if consonants > 2 and duplicates != consonants - 1:
result += term + " " + w
consonants = 1
duplicates = 0
term = ""
else:
term += w
result += term
if result.startswith(" "):
result == result[1:]
print(result) | ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR NUMBER VAR VAR VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR STRING VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING VAR VAR VAR VAR IF FUNC_CALL VAR STRING EXPR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = list(input())
h = "aeuio "
l = ""
i = 0
while i < len(s) - 2:
a = s[i]
b = s[i + 1]
c = s[i + 2]
if h.find(a) != -1 or h.find(b) != -1 or h.find(c) != -1 or a == b and a == c:
i += 1
else:
s.insert(i + 2, " ")
i += 1
for i in range(len(s)):
l += s[i]
print(l) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER STRING VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
j = 0
a = ["a", "e", "i", "o", "u"]
for i in range(len(s)):
if j > 1:
if not s[i] in a:
if (
not s[i - 1] in a
and not s[i - 2] in a
and not s[i] == s[i - 1] == s[i - 2]
):
print(" ", end="")
j = 0
j += 1
print(s[i], end="") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST STRING STRING STRING STRING STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING STRING ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
n = len(s)
v = ["a", "e", "i", "o", "u"]
l = []
i = 0
j = 0
while i < n - 1:
if s[i] not in v and s[i + 1] not in v:
c = 1
x = set(s[i])
while s[i] not in v and s[i + 1] not in v:
if c >= 2:
if s[i] == s[i + 1] and len(x) == 1:
pass
else:
l.append(s[j : i + 1])
j = i + 1
i += 1
break
c += 1
x.add(s[i + 1])
i += 1
if i == n - 1:
break
else:
i += 1
if j != n:
l.append(s[j:])
print(" ".join(l)) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR WHILE VAR VAR VAR VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | t = y = ""
k = s = 0
for x in input():
if x in "aeiou":
t += x
k = s = 0
elif k == 1:
if x == y:
t += x
s = 1
elif s:
t += " " + x
k, s, y = 1, 0, x
else:
t += x
k = 2
elif k == 2:
t += " " + x
k, s, y = 1, 0, x
else:
t += x
k, y = 1, x
print(t) | ASSIGN VAR VAR STRING ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR IF VAR STRING VAR VAR ASSIGN VAR VAR NUMBER IF VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR BIN_OP STRING VAR ASSIGN VAR VAR VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP STRING VAR ASSIGN VAR VAR VAR NUMBER NUMBER VAR VAR VAR ASSIGN VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | a = input()
gl = set(["a", "e", "i", "o", "u"])
cn = [0] * (200 + 1)
ans = []
for i in range(len(a)):
if a[i] in gl:
for i1 in range(ord("a"), ord("z") + 1):
cn[i1] = 0
continue
cn[ord(a[i])] += 1
st = set()
cnt = 0
for j in range(ord("a"), ord("z") + 1):
if chr(j) in gl or cn[j] == 0:
continue
cnt += cn[j]
st.add(j)
if len(st) > 1 and cnt >= 3:
ans.append(i)
for i1 in range(ord("a"), ord("z") + 1):
cn[i1] = 0
cn[ord(a[i])] += 1
break
for i in range(len(a)):
if i in ans:
print(" ", end="")
print(a[i], end="") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR BIN_OP LIST NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER ASSIGN VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER ASSIGN VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING STRING EXPR FUNC_CALL VAR VAR VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | vowels = {"a", "e", "i", "o", "u"}
s = input()
count_consonant = 0
exists_different = False
start = 0
for i in range(0, len(s)):
if s[i] not in vowels:
exists_different = count_consonant >= 1 and (
exists_different or s[i] != s[i - 1]
)
count_consonant += 1
if count_consonant >= 3 and exists_different:
print(s[start:i], end=" ")
start = i
count_consonant = 1
exists_different = False
else:
count_consonant = 0
exists_different = False
print(s[start : len(s)]) | ASSIGN VAR STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR STRING ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | data = ["a", "e", "i", "o", "u"]
s = list(input())
now = 0
i = 0
while i < len(s):
if s[i] not in data:
now += 1
else:
now = 0
if now >= 3:
if s[i] == s[i - 1] == s[i - 2]:
pass
else:
s.insert(i, " ")
i += 1
now = 1
i += 1
print("".join(map(str, s))) | ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
def ans(s):
vowel = {"a", "e", "i", "o", "u"}
cc = 0
begin = 0
for i in range(len(s)):
if s[i] in vowel:
cc = 0
else:
cc += 1
if cc >= 3 and (s[i] != s[i - 1] or s[i] != s[i - 2] or s[i - 1] != s[i - 2]):
print(s[begin:i], end=" ")
cc = 1
begin = i
print(s[begin:])
ans(s) | ASSIGN VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | ch = str(input())
i = 0
C = ["a", "e", "i", "u", "o"]
if len(ch) <= 2:
print(ch)
else:
while i <= len(ch) - 3:
if (
not ch[i] in C
and not ch[i + 1] in C
and not ch[i + 2] in C
and (ch[i] != ch[i + 1] or ch[i] != ch[i + 2] or ch[i + 1] != ch[i + 2])
):
ch = ch[0 : i + 2] + " " + ch[i + 2 : len(ch)]
i = i + 3
else:
i = i + 1
print(ch) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST STRING STRING STRING STRING STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | a = str(input())
sgl = "bcdfghjklmnpqrstvwxyz"
ans = ""
for i in range(len(a)):
ans += a[i]
if len(ans) > 2:
if (
sgl.count(ans[-1])
and sgl.count(ans[-2])
and sgl.count(ans[-3])
and (ans[-1] != ans[-2] or ans[-1] != ans[-3])
):
ans = ans[:-1] + " " + ans[-1]
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER STRING VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | gl = ["a", "e", "i", "o", "u"]
def mist(a, b, c):
if a in gl or b in gl or c in gl:
return False
if a == b and b == c:
return False
return True
s = input()
n = len(s)
if n == 1:
print(s[0])
else:
print(s[0] + s[1], end="")
last1 = s[0]
last2 = s[1]
for i in range(2, n):
now = s[i]
if now in gl:
print(now, end="")
last1, last2 = last2, now
continue
if mist(last1, last2, now):
print(" " + now, end="")
last1 = "a"
last2 = now
else:
print(now, end="")
last1, last2 = last2, now | ASSIGN VAR LIST STRING STRING STRING STRING STRING FUNC_DEF IF VAR VAR VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER STRING ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR STRING ASSIGN VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP STRING VAR STRING ASSIGN VAR STRING ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR STRING ASSIGN VAR VAR VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
vowels = "a", "e", "u", "o", "i"
res = ""
i = 0
while i < len(s) - 2:
if (
s[i] not in vowels
and s[i + 1] not in vowels
and s[i + 2] not in vowels
and s[i : i + 3] != s[i] * 3
):
res += s[i : i + 2] + " "
i += 2
else:
res += s[i]
i += 1
res += s[i:]
print(res) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING STRING STRING STRING STRING ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR VAR VAR VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | def main():
str = input() + " "
vows = {"a", "e", "i", "o", "u"}
idx = 0
res = []
while idx + 2 < len(str):
if (
str[idx] not in vows
and str[idx + 1] not in vows
and str[idx + 2] not in vows
and not str[idx] == str[idx + 1] == str[idx + 2]
):
res.append(str[idx : idx + 2])
res.append(" ")
idx += 2
else:
res.append(str[idx])
idx += 1
print("".join(res).rstrip())
main() | FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR STRING ASSIGN VAR STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL STRING VAR EXPR FUNC_CALL VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | vowels = "aeiou"
w = input()
letter = "a"
count = 0
single = True
for c in w:
if c in vowels:
letter = c
count = 0
single = True
print(c, end="")
else:
if count > 0 and c != letter:
single = False
count += 1
letter = c
if count >= 3 and not single:
count = 1
single = True
print(" " + c, end="")
else:
print(c, end="")
print() | ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR STRING IF VAR NUMBER VAR VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP STRING VAR STRING EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
gl = ["a", "e", "i", "o", "u"]
uk = 0
otv = ""
for i in range(len(s)):
if s[i] in gl:
uk = 0
otv += s[i]
else:
uk += 1
if uk >= 3 and (s[i] != s[i - 1] or s[i] != s[i - 2]):
uk = 1
otv += " " + s[i]
else:
otv += s[i]
print(otv) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP STRING VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
letters = ["a", "e", "i", "o", "u"]
index = []
i = 0
while i < len(s) - 2:
x = s[i] == s[i + 1] == s[i + 2]
if (
s[i] not in letters
and s[i + 1] not in letters
and s[i + 2] not in letters
and x == False
):
index.append(i + 1)
i += 2
else:
i += 1
if len(index) != 0:
s1 = s[: index[0] + 1] + " " + s[index[0] + 1 :]
space = 1
index_1 = index[1:]
for i in index_1:
s1 = s1[: i + 1 + space] + " " + s[i + 1 :]
space += 1
print(s1)
else:
print(s) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER STRING VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR STRING VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | vowels = {"a", "e", "i", "o", "u"}
s = input()
n = len(s)
cur = []
ans = []
for c in s:
if len(cur) < 2:
cur.append(c)
elif (
c not in vowels
and cur[-1] not in vowels
and cur[-2] not in vowels
and len(set([c, cur[-1], cur[-2]])) != 1
):
ans.append("".join(cur))
cur = [c]
else:
cur.append(c)
if len(cur):
ans.append("".join(cur))
print(" ".join(ans)) | ASSIGN VAR STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR LIST VAR VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR LIST VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
s2 = ""
alp = "aeiou"
i = 0
flag = 0
while i < len(s):
if alp.find(s[i]) != -1:
s2 += s[i]
i += 1
flag = 0
else:
flag += 1
if flag == 3:
if s[i - 2] == s[i - 1] and s[i] == s[i - 1]:
s2 += s[i]
flag -= 1
i += 1
else:
s2 += " "
flag = 0
else:
s2 += s[i]
i += 1
print(s2) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER VAR STRING ASSIGN VAR NUMBER VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
glas = "aeiou"
i = 1
while i < len(s) - 1:
if s[i - 1] not in glas and s[i] not in glas and s[i + 1] not in glas:
if s[i - 1] != s[i] or s[i] != s[i + 1] or s[i - 1] != s[i + 1]:
s = s[: i + 1] + " " + s[i + 1 :]
i += 3
continue
i += 1
print(s) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
d = s[0]
a = ["a", "e", "i", "o", "u"]
i = 1
while i < len(s) - 1:
d += s[i]
if (
(not s[i] in a)
& (not s[i + 1] in a)
& (not s[i - 1] in a)
& (not s[i] == s[i + 1] == s[i - 1])
):
d += " " + s[i + 1]
i += 1
i += 1
if i < len(s):
d += s[i]
print(d) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR IF BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP STRING VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | string = input()
s = "ywrtpsdfqghjklzxcvbnm"
posl = 0
word_answer = ""
otvet = 0
for i in range(len(string)):
if s.find(string[i]) != -1:
otvet += 1
else:
otvet = 0
if otvet >= 3:
if (
string[i] != string[i - 1]
or string[i] != string[i - 2]
or string[i - 1] != string[i - 2]
):
word_answer = word_answer + string[posl:i] + " "
posl = i
otvet = 0
if s.find(string[i]) != -1:
otvet += 1
word_answer = word_answer + string[posl:]
print(word_answer) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR STRING ASSIGN VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | x = str(input())
n, count, count1, ans = len(x), 0, 0, [0] * 3001
for i in range(n):
if x[i] in {"a", "e", "i", "o", "u"}:
count = 0
else:
count += 1
if count == 3:
if x[i] == x[i - 1] == x[i - 2]:
count = 2
else:
ans[i - 1] = 1
count = 1
for i in range(n):
print(x[i], end="")
if ans[i] == 1:
print(" ", end="") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING STRING STRING STRING STRING ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING IF VAR VAR NUMBER EXPR FUNC_CALL VAR STRING STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
g = "aieou"
f = 0
k = 0
p = ""
news = ""
for i in range(len(s)):
if s[i] not in g:
k += 1
if k >= 3 and s[i] != p[-1]:
news = news + " " + s[i]
k = 1
p = s[i]
elif k >= 3 and s[i] == p[-1]:
if len(p) >= 2:
if s[i] == p[0]:
news += s[i]
p += s[i]
else:
news = news + " " + s[i]
p = s[i]
k = 1
else:
p += s[i]
news += s[i]
else:
news += s[i]
k = 0
p = ""
print(news) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER IF VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR STRING VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR STRING VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | def check(s):
if s[0] == s[1] == s[2]:
return False
return True
s = input()
gl = ["a", "e", "i", "o", "u"]
cnt = 0
last_letter = ""
i = 0
while i < len(s):
if s[i] not in gl:
cnt += 1
else:
cnt = 0
if cnt == 3:
if check(s[i - 2 : i + 1]):
s = s[:i] + " " + s[i:]
cnt = 1
i += 1
else:
cnt -= 1
i += 1
print(s) | FUNC_DEF IF VAR NUMBER VAR NUMBER VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR STRING VAR VAR ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
s2 = s[0]
gl = "aeiou"
count = 0
mark = 0
if s[0] not in gl:
count = 1
for i in range(1, len(s)):
if s[i] not in gl:
if count >= 2:
if mark == 1 or s[i] != s[i - 1]:
s2 += " "
count = 1
mark = 0
else:
if s[i] != s[i - 1] and s[i - 1] not in gl:
mark = 1
count += 1
else:
count = 0
mark = 0
s2 += s[i]
print(s2) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR IF VAR NUMBER IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | word = input()
output = ""
v = ["a", "e", "i", "o", "u"]
s = 0
while True:
if s + 1 < len(word) and s + 2 < len(word):
if (
v.count(word[s]) == 0
and v.count(word[s + 1]) == 0
and v.count(word[s + 2]) == 0
):
if len(set(word[s] + word[s + 1] + word[s + 2])) > 1:
output += word[s] + word[s + 1] + " "
s += 2
else:
output += word[s]
s += 1
else:
output += word[s]
s += 1
else:
output += word[s]
s += 1
if s >= len(word):
break
print(output) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR NUMBER WHILE NUMBER IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
c = 0
d = 0
p = ""
for i in s:
if i == "a" or i == "e" or i == "i" or i == "o" or i == "u":
c = 0
d = 0
p = ""
else:
c += 1
if c > 1:
if p != "" and i not in p:
d = 1
p += i
if c >= 3 and d:
print(" ", end="")
c = 1
d = 0
p = i
print(i, end="") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR VAR IF VAR STRING VAR STRING VAR STRING VAR STRING VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING VAR NUMBER IF VAR NUMBER IF VAR STRING VAR VAR ASSIGN VAR NUMBER VAR VAR IF VAR NUMBER VAR EXPR FUNC_CALL VAR STRING STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | string = input()
l = len(string)
string += " "
consonents = "bcdfghjklmnpqrstvwxyz"
vowels = "aeiou"
output = ""
if len(string) < 3:
print(string)
else:
i = 0
while i < l:
if string[i] in vowels or string[i + 1] in vowels or string[i + 2] in vowels:
output += string[i]
i += 1
elif string[i] == string[i + 1] == string[i + 2]:
output += string[i]
i += 1
elif (
string[i] != string[i + 1]
or string[i + 1] != string[i + 2]
or string[i] != string[i + 2]
or string[i] != string[i + 1] != string[i + 2]
):
output += string[i] + string[i + 1] + " "
i += 2
print(output.strip()) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER STRING VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | s = input()
fl = False
l1 = ""
l2 = ""
l = 0
gl = {"a", "e", "i", "o", "u"}
for i in range(len(s)):
if s[i] in gl:
l1 = ""
l2 = ""
elif l2 == "":
l2 = s[i]
continue
elif l1 == "":
l1 = s[i]
continue
elif s[i] == l2 and s[i] == l1:
continue
else:
print(s[l:i] + " ", end="")
l1 = ""
l2 = s[i]
l = i
print(s[l:]) | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR STRING STRING STRING STRING STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR VAR VAR IF VAR STRING ASSIGN VAR VAR VAR IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR STRING STRING ASSIGN VAR STRING ASSIGN VAR VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | string = input()
def sogl(ch):
if ch != "a" and ch != "e" and ch != "i" and ch != "o" and ch != "u":
return True
return False
fast = 0
for i in range(len(string)):
if fast > 0:
fast -= 1
continue
if (
i < len(string) - 2
and not (string[i] == string[i + 1] and string[i] == string[i + 2])
and sogl(string[i])
and sogl(string[i + 1])
and sogl(string[i + 2])
):
print(string[i] + string[i + 1] + " ", end="")
fast = 1
continue
print(string[i], end="") | ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF VAR STRING VAR STRING VAR STRING VAR STRING VAR STRING RETURN NUMBER RETURN NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER STRING STRING ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | symbol = ["a", "e", "i", "o", "u"]
a = input()
i = 2
while i < len(a):
if (
not a[i] in symbol
and not a[i - 1] in symbol
and not a[i - 2] in symbol
and a[i - 2 : i + 1] != a[i - 2] * 3
):
a = a[:i] + " " + a[i:]
i = i + 3
else:
i = i + 1
print(a) | ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR STRING VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
* the following words have typos: "hellno", "hackcerrs" and "backtothefutttture";
* the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Examples
Input
hellno
Output
hell no
Input
abacaba
Output
abacaba
Input
asdfasdf
Output
asd fasd f | orph = ["a", "e", "i", "o", "u"]
string = input()
ctr = 0
out = ""
for i, c in enumerate(string):
if c in orph:
ctr = 0
else:
ctr += 1
if ctr > 2:
if not (c == string[i - 1] and c == string[i - 2]):
out = out + " "
ctr = 1
out = out + c
print(out) | ASSIGN VAR LIST STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR STRING ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array of integers $a_1, a_2, \ldots, a_n$ and an integer $x$.
You need to select the maximum number of elements in the array, such that for every subsegment $a_l, a_{l + 1}, \ldots, a_r$ containing strictly more than one element $(l < r)$, either:
At least one element on this subsegment is not selected, or
$a_l + a_{l+1} + \ldots + a_r \geq x \cdot (r - l + 1)$.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 10$): the number of test cases.
The descriptions of $t$ test cases follow, three lines per test case.
In the first line you are given one integer $n$ ($1 \leq n \leq 50000$): the number of integers in the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-100000 \leq a_i \leq 100000$).
The third line contains one integer $x$ ($-100000 \leq x \leq 100000$).
-----Output-----
For each test case, print one integer: the maximum number of elements that you can select.
-----Examples-----
Input
4
5
1 2 3 4 5
2
10
2 4 2 4 2 4 2 4 2 4
3
3
-10 -5 -10
-8
3
9 9 -3
5
Output
4
8
2
2
-----Note-----
In the first example, one valid way to select the elements is $[\underline{1}, 2, \underline{3}, \underline{4}, \underline{5}]$. All subsegments satisfy at least one of the criteria. For example, for the subsegment $l = 1$, $r = 2$ we have that the element $2$ is not selected, satisfying the first criterion. For the subsegment $l = 3$, $r = 5$ we have $3 + 4 + 5 = 12 \ge 2 \cdot 3$, satisfying the second criterion.
We can't select all elements, because in this case for $l = 1$, $r = 2$ all elements are selected and we have $a_1 + a_2 = 3 < 2 \cdot 2$. Thus, the maximum number of selected elements is $4$.
In the second example, one valid solution is $[\underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}]$.
In the third example, one valid solution is $[\underline{-10}, -5, \underline{-10}]$.
In the fourth example, one valid solution is $[\underline{9}, \underline{9}, -3]$. | def Rep_Edges(n, l, x):
i = 0
while i < n:
l[i] = l[i] - x
i = i + 1
ans = n
j = -1
for i in range(1, n):
if (i - 1 > j) & (l[i] + l[i - 1] < 0):
ans = ans - 1
j = i
elif (i - 2 > j) & (l[i] + l[i - 1] + l[i - 2] < 0):
ans = ans - 1
j = i
return ans
t = int(input())
ls = []
ln = []
la = []
lx = []
for i in range(t):
n = int(input())
s = str(input())
x = int(input())
ln.append(n)
ls.append(s)
lx.append(x)
for i in range(len(ls)):
n = ln[i]
s = ls[i]
x = lx[i]
l = []
for j in s.split(" "):
l.append(int(j))
ans = Rep_Edges(n, l, x)
la.append(ans)
for a in la:
print(a) | FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR IF BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array of integers $a_1, a_2, \ldots, a_n$ and an integer $x$.
You need to select the maximum number of elements in the array, such that for every subsegment $a_l, a_{l + 1}, \ldots, a_r$ containing strictly more than one element $(l < r)$, either:
At least one element on this subsegment is not selected, or
$a_l + a_{l+1} + \ldots + a_r \geq x \cdot (r - l + 1)$.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 10$): the number of test cases.
The descriptions of $t$ test cases follow, three lines per test case.
In the first line you are given one integer $n$ ($1 \leq n \leq 50000$): the number of integers in the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-100000 \leq a_i \leq 100000$).
The third line contains one integer $x$ ($-100000 \leq x \leq 100000$).
-----Output-----
For each test case, print one integer: the maximum number of elements that you can select.
-----Examples-----
Input
4
5
1 2 3 4 5
2
10
2 4 2 4 2 4 2 4 2 4
3
3
-10 -5 -10
-8
3
9 9 -3
5
Output
4
8
2
2
-----Note-----
In the first example, one valid way to select the elements is $[\underline{1}, 2, \underline{3}, \underline{4}, \underline{5}]$. All subsegments satisfy at least one of the criteria. For example, for the subsegment $l = 1$, $r = 2$ we have that the element $2$ is not selected, satisfying the first criterion. For the subsegment $l = 3$, $r = 5$ we have $3 + 4 + 5 = 12 \ge 2 \cdot 3$, satisfying the second criterion.
We can't select all elements, because in this case for $l = 1$, $r = 2$ all elements are selected and we have $a_1 + a_2 = 3 < 2 \cdot 2$. Thus, the maximum number of selected elements is $4$.
In the second example, one valid solution is $[\underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}]$.
In the third example, one valid solution is $[\underline{-10}, -5, \underline{-10}]$.
In the fourth example, one valid solution is $[\underline{9}, \underline{9}, -3]$. | def solver():
n = int(input())
for _ in range(n):
len_input = int(input())
(*array,) = map(int, input().split())
x = int(input())
incriminated = set()
for i in range(1, len_input):
if i == 1:
if array[i - 1] + array[i] < x * 2:
incriminated.add(i)
elif i - 2 in incriminated:
if i - 1 in incriminated:
continue
elif array[i - 1] + array[i] < 2 * x:
incriminated.add(i)
elif not i - 1 in incriminated:
if array[i - 2] + array[i - 1] + array[i] < 3 * x:
incriminated.add(i)
if array[i - 1] + array[i] < 2 * x:
incriminated.add(i)
print(len_input - len(incriminated))
solver() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR NUMBER IF BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER VAR IF BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR IF BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given an array of integers $a_1, a_2, \ldots, a_n$ and an integer $x$.
You need to select the maximum number of elements in the array, such that for every subsegment $a_l, a_{l + 1}, \ldots, a_r$ containing strictly more than one element $(l < r)$, either:
At least one element on this subsegment is not selected, or
$a_l + a_{l+1} + \ldots + a_r \geq x \cdot (r - l + 1)$.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 10$): the number of test cases.
The descriptions of $t$ test cases follow, three lines per test case.
In the first line you are given one integer $n$ ($1 \leq n \leq 50000$): the number of integers in the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-100000 \leq a_i \leq 100000$).
The third line contains one integer $x$ ($-100000 \leq x \leq 100000$).
-----Output-----
For each test case, print one integer: the maximum number of elements that you can select.
-----Examples-----
Input
4
5
1 2 3 4 5
2
10
2 4 2 4 2 4 2 4 2 4
3
3
-10 -5 -10
-8
3
9 9 -3
5
Output
4
8
2
2
-----Note-----
In the first example, one valid way to select the elements is $[\underline{1}, 2, \underline{3}, \underline{4}, \underline{5}]$. All subsegments satisfy at least one of the criteria. For example, for the subsegment $l = 1$, $r = 2$ we have that the element $2$ is not selected, satisfying the first criterion. For the subsegment $l = 3$, $r = 5$ we have $3 + 4 + 5 = 12 \ge 2 \cdot 3$, satisfying the second criterion.
We can't select all elements, because in this case for $l = 1$, $r = 2$ all elements are selected and we have $a_1 + a_2 = 3 < 2 \cdot 2$. Thus, the maximum number of selected elements is $4$.
In the second example, one valid solution is $[\underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}]$.
In the third example, one valid solution is $[\underline{-10}, -5, \underline{-10}]$.
In the fourth example, one valid solution is $[\underline{9}, \underline{9}, -3]$. | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
x = int(input())
ans = 0
for i in range(n):
a[i] -= x
sa = [a[0]]
for i in range(1, n):
sa.append(sa[-1] + a[i])
i = 1
l = a[0]
while i < n:
if l + a[i] < 0:
ans += 1
i += 1
if i < n:
l = a[i]
else:
l = min(a[i], l + a[i])
i += 1
print(n - ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR LIST VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER WHILE VAR VAR IF BIN_OP VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array of integers $a_1, a_2, \ldots, a_n$ and an integer $x$.
You need to select the maximum number of elements in the array, such that for every subsegment $a_l, a_{l + 1}, \ldots, a_r$ containing strictly more than one element $(l < r)$, either:
At least one element on this subsegment is not selected, or
$a_l + a_{l+1} + \ldots + a_r \geq x \cdot (r - l + 1)$.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 10$): the number of test cases.
The descriptions of $t$ test cases follow, three lines per test case.
In the first line you are given one integer $n$ ($1 \leq n \leq 50000$): the number of integers in the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-100000 \leq a_i \leq 100000$).
The third line contains one integer $x$ ($-100000 \leq x \leq 100000$).
-----Output-----
For each test case, print one integer: the maximum number of elements that you can select.
-----Examples-----
Input
4
5
1 2 3 4 5
2
10
2 4 2 4 2 4 2 4 2 4
3
3
-10 -5 -10
-8
3
9 9 -3
5
Output
4
8
2
2
-----Note-----
In the first example, one valid way to select the elements is $[\underline{1}, 2, \underline{3}, \underline{4}, \underline{5}]$. All subsegments satisfy at least one of the criteria. For example, for the subsegment $l = 1$, $r = 2$ we have that the element $2$ is not selected, satisfying the first criterion. For the subsegment $l = 3$, $r = 5$ we have $3 + 4 + 5 = 12 \ge 2 \cdot 3$, satisfying the second criterion.
We can't select all elements, because in this case for $l = 1$, $r = 2$ all elements are selected and we have $a_1 + a_2 = 3 < 2 \cdot 2$. Thus, the maximum number of selected elements is $4$.
In the second example, one valid solution is $[\underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}]$.
In the third example, one valid solution is $[\underline{-10}, -5, \underline{-10}]$.
In the fourth example, one valid solution is $[\underline{9}, \underline{9}, -3]$. | tt = int(input())
for TC in range(tt):
n = int(input())
a = [int(x) for x in input().split()]
x = int(input())
for i in range(n):
a[i] -= x
v = []
for i in range(1, n):
if a[i] + a[i - 1] < 0:
v.append((i, 2))
for i in range(2, n):
if (
a[i] + a[i - 1] >= 0
and a[i - 1] + a[i - 2] >= 0
and a[i] + a[i - 1] + a[i - 2] < 0
):
v.append((i, 3))
v.sort()
if len(v) == 0:
print(n)
continue
deleted = 0
cnt = 1
for i in range(1, len(v)):
if v[i][0] - v[i][1] + 1 == v[i - 1][0]:
cnt += 1
else:
deleted += (cnt + 1) // 2
cnt = 1
deleted += (cnt + 1) // 2
print(n - deleted) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array of integers $a_1, a_2, \ldots, a_n$ and an integer $x$.
You need to select the maximum number of elements in the array, such that for every subsegment $a_l, a_{l + 1}, \ldots, a_r$ containing strictly more than one element $(l < r)$, either:
At least one element on this subsegment is not selected, or
$a_l + a_{l+1} + \ldots + a_r \geq x \cdot (r - l + 1)$.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 10$): the number of test cases.
The descriptions of $t$ test cases follow, three lines per test case.
In the first line you are given one integer $n$ ($1 \leq n \leq 50000$): the number of integers in the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-100000 \leq a_i \leq 100000$).
The third line contains one integer $x$ ($-100000 \leq x \leq 100000$).
-----Output-----
For each test case, print one integer: the maximum number of elements that you can select.
-----Examples-----
Input
4
5
1 2 3 4 5
2
10
2 4 2 4 2 4 2 4 2 4
3
3
-10 -5 -10
-8
3
9 9 -3
5
Output
4
8
2
2
-----Note-----
In the first example, one valid way to select the elements is $[\underline{1}, 2, \underline{3}, \underline{4}, \underline{5}]$. All subsegments satisfy at least one of the criteria. For example, for the subsegment $l = 1$, $r = 2$ we have that the element $2$ is not selected, satisfying the first criterion. For the subsegment $l = 3$, $r = 5$ we have $3 + 4 + 5 = 12 \ge 2 \cdot 3$, satisfying the second criterion.
We can't select all elements, because in this case for $l = 1$, $r = 2$ all elements are selected and we have $a_1 + a_2 = 3 < 2 \cdot 2$. Thus, the maximum number of selected elements is $4$.
In the second example, one valid solution is $[\underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}]$.
In the third example, one valid solution is $[\underline{-10}, -5, \underline{-10}]$.
In the fourth example, one valid solution is $[\underline{9}, \underline{9}, -3]$. | D = input
C = int
for H in range(C(D())):
I = C(D())
B = list(map(C, D().split()))
G = C(D())
B = [(A - G) for A in B]
A = 10**18
E = 0
for F in B:
A += F
if A < 0:
A = 10**18
else:
E += 1
A = min(A, F)
print(E) | ASSIGN VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array of integers $a_1, a_2, \ldots, a_n$ and an integer $x$.
You need to select the maximum number of elements in the array, such that for every subsegment $a_l, a_{l + 1}, \ldots, a_r$ containing strictly more than one element $(l < r)$, either:
At least one element on this subsegment is not selected, or
$a_l + a_{l+1} + \ldots + a_r \geq x \cdot (r - l + 1)$.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 10$): the number of test cases.
The descriptions of $t$ test cases follow, three lines per test case.
In the first line you are given one integer $n$ ($1 \leq n \leq 50000$): the number of integers in the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-100000 \leq a_i \leq 100000$).
The third line contains one integer $x$ ($-100000 \leq x \leq 100000$).
-----Output-----
For each test case, print one integer: the maximum number of elements that you can select.
-----Examples-----
Input
4
5
1 2 3 4 5
2
10
2 4 2 4 2 4 2 4 2 4
3
3
-10 -5 -10
-8
3
9 9 -3
5
Output
4
8
2
2
-----Note-----
In the first example, one valid way to select the elements is $[\underline{1}, 2, \underline{3}, \underline{4}, \underline{5}]$. All subsegments satisfy at least one of the criteria. For example, for the subsegment $l = 1$, $r = 2$ we have that the element $2$ is not selected, satisfying the first criterion. For the subsegment $l = 3$, $r = 5$ we have $3 + 4 + 5 = 12 \ge 2 \cdot 3$, satisfying the second criterion.
We can't select all elements, because in this case for $l = 1$, $r = 2$ all elements are selected and we have $a_1 + a_2 = 3 < 2 \cdot 2$. Thus, the maximum number of selected elements is $4$.
In the second example, one valid solution is $[\underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}]$.
In the third example, one valid solution is $[\underline{-10}, -5, \underline{-10}]$.
In the fourth example, one valid solution is $[\underline{9}, \underline{9}, -3]$. | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
x = int(input())
ans = 0
Sum = 2e18
for i in a:
Sum += i - x
if Sum < 0:
Sum = 2e18
else:
Sum = min(Sum, i - x)
ans += 1
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given an array of integers $a_1, a_2, \ldots, a_n$ and an integer $x$.
You need to select the maximum number of elements in the array, such that for every subsegment $a_l, a_{l + 1}, \ldots, a_r$ containing strictly more than one element $(l < r)$, either:
At least one element on this subsegment is not selected, or
$a_l + a_{l+1} + \ldots + a_r \geq x \cdot (r - l + 1)$.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 10$): the number of test cases.
The descriptions of $t$ test cases follow, three lines per test case.
In the first line you are given one integer $n$ ($1 \leq n \leq 50000$): the number of integers in the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-100000 \leq a_i \leq 100000$).
The third line contains one integer $x$ ($-100000 \leq x \leq 100000$).
-----Output-----
For each test case, print one integer: the maximum number of elements that you can select.
-----Examples-----
Input
4
5
1 2 3 4 5
2
10
2 4 2 4 2 4 2 4 2 4
3
3
-10 -5 -10
-8
3
9 9 -3
5
Output
4
8
2
2
-----Note-----
In the first example, one valid way to select the elements is $[\underline{1}, 2, \underline{3}, \underline{4}, \underline{5}]$. All subsegments satisfy at least one of the criteria. For example, for the subsegment $l = 1$, $r = 2$ we have that the element $2$ is not selected, satisfying the first criterion. For the subsegment $l = 3$, $r = 5$ we have $3 + 4 + 5 = 12 \ge 2 \cdot 3$, satisfying the second criterion.
We can't select all elements, because in this case for $l = 1$, $r = 2$ all elements are selected and we have $a_1 + a_2 = 3 < 2 \cdot 2$. Thus, the maximum number of selected elements is $4$.
In the second example, one valid solution is $[\underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}]$.
In the third example, one valid solution is $[\underline{-10}, -5, \underline{-10}]$.
In the fourth example, one valid solution is $[\underline{9}, \underline{9}, -3]$. | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
x = int(input())
a = [(c - x) for c in a]
begin = True
unchosen = 0
sum = 0
for i in range(n):
if begin:
sum = a[i]
begin = False
else:
sum += a[i]
if sum < 0:
begin = True
unchosen += 1
else:
sum = min(sum, a[i])
print(n - unchosen) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array of integers $a_1, a_2, \ldots, a_n$ and an integer $x$.
You need to select the maximum number of elements in the array, such that for every subsegment $a_l, a_{l + 1}, \ldots, a_r$ containing strictly more than one element $(l < r)$, either:
At least one element on this subsegment is not selected, or
$a_l + a_{l+1} + \ldots + a_r \geq x \cdot (r - l + 1)$.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 10$): the number of test cases.
The descriptions of $t$ test cases follow, three lines per test case.
In the first line you are given one integer $n$ ($1 \leq n \leq 50000$): the number of integers in the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-100000 \leq a_i \leq 100000$).
The third line contains one integer $x$ ($-100000 \leq x \leq 100000$).
-----Output-----
For each test case, print one integer: the maximum number of elements that you can select.
-----Examples-----
Input
4
5
1 2 3 4 5
2
10
2 4 2 4 2 4 2 4 2 4
3
3
-10 -5 -10
-8
3
9 9 -3
5
Output
4
8
2
2
-----Note-----
In the first example, one valid way to select the elements is $[\underline{1}, 2, \underline{3}, \underline{4}, \underline{5}]$. All subsegments satisfy at least one of the criteria. For example, for the subsegment $l = 1$, $r = 2$ we have that the element $2$ is not selected, satisfying the first criterion. For the subsegment $l = 3$, $r = 5$ we have $3 + 4 + 5 = 12 \ge 2 \cdot 3$, satisfying the second criterion.
We can't select all elements, because in this case for $l = 1$, $r = 2$ all elements are selected and we have $a_1 + a_2 = 3 < 2 \cdot 2$. Thus, the maximum number of selected elements is $4$.
In the second example, one valid solution is $[\underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}]$.
In the third example, one valid solution is $[\underline{-10}, -5, \underline{-10}]$.
In the fourth example, one valid solution is $[\underline{9}, \underline{9}, -3]$. | t = int(input())
for i in range(t):
n = int(input())
l = list(map(int, input().split()))
x = int(input())
for i in range(n):
l[i] -= x
l1 = [0]
m = 0
for i in range(n):
m += l[i]
l1.append(m)
high = -1000000
high1 = -1000000
h1 = -1
m = 0
cur = 0
for i in range(n):
if i - 1 >= cur:
if l[i] + l[i - 1] < 0:
cur = i + 1
m += 1
if i - 2 >= cur:
if l[i] + l[i - 1] + l[i - 2] < 0:
cur = i + 1
m += 1
print(n - m) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR IF BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR IF BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
You are given an array of integers $a_1, a_2, \ldots, a_n$ and an integer $x$.
You need to select the maximum number of elements in the array, such that for every subsegment $a_l, a_{l + 1}, \ldots, a_r$ containing strictly more than one element $(l < r)$, either:
At least one element on this subsegment is not selected, or
$a_l + a_{l+1} + \ldots + a_r \geq x \cdot (r - l + 1)$.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 10$): the number of test cases.
The descriptions of $t$ test cases follow, three lines per test case.
In the first line you are given one integer $n$ ($1 \leq n \leq 50000$): the number of integers in the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($-100000 \leq a_i \leq 100000$).
The third line contains one integer $x$ ($-100000 \leq x \leq 100000$).
-----Output-----
For each test case, print one integer: the maximum number of elements that you can select.
-----Examples-----
Input
4
5
1 2 3 4 5
2
10
2 4 2 4 2 4 2 4 2 4
3
3
-10 -5 -10
-8
3
9 9 -3
5
Output
4
8
2
2
-----Note-----
In the first example, one valid way to select the elements is $[\underline{1}, 2, \underline{3}, \underline{4}, \underline{5}]$. All subsegments satisfy at least one of the criteria. For example, for the subsegment $l = 1$, $r = 2$ we have that the element $2$ is not selected, satisfying the first criterion. For the subsegment $l = 3$, $r = 5$ we have $3 + 4 + 5 = 12 \ge 2 \cdot 3$, satisfying the second criterion.
We can't select all elements, because in this case for $l = 1$, $r = 2$ all elements are selected and we have $a_1 + a_2 = 3 < 2 \cdot 2$. Thus, the maximum number of selected elements is $4$.
In the second example, one valid solution is $[\underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}, 2, \underline{4}, \underline{2}, \underline{4}]$.
In the third example, one valid solution is $[\underline{-10}, -5, \underline{-10}]$.
In the fourth example, one valid solution is $[\underline{9}, \underline{9}, -3]$. | import sys
input = sys.stdin.readline
def ri():
return int(input())
def rf():
return list(map(float, input().split()))
def rl():
return list(map(int, input().split()))
def rs():
return input().rstrip()
def out_list(x):
return " ".join(map(str, x))
def solve_case():
n = ri()
A = rl()
x = ri()
res = 0
j = 0
for i in range(n):
ok = True
if i - j >= 1 and A[i - 1] + A[i] < 2 * x:
ok = False
if i - j >= 2 and A[i - 2] + A[i - 1] + A[i] < 3 * x:
ok = False
if not ok:
res += 1
j = i + 1
print(n - res)
T = ri()
for ti in range(1, T + 1):
solve_case() | IMPORT ASSIGN VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL STRING FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | def fine():
f, c, n = 1, 1, 1
yield 0
while True:
yield f
n += 1
c = c * (4 * n - 6) // n
f = (c - f) // 2
f = fine()
n = int(input())
print((sum(next(f) for _ in range(n + 3)) - 1) % (10**9 + 7)) | FUNC_DEF ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER EXPR NUMBER WHILE NUMBER EXPR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | n = int(input())
dp = [[([0] * (n + 1)) for i in range(n + 1)] for j in range(2)]
dp[0][0][0] = 0
dp[1][0][0] = 0
mod = 10**9 + 7
for i in range(n + 1):
for j in range(i, n + 1):
if i == 0 and j == 0:
continue
dp[0][i][j] = (dp[1][i - 1][j] + dp[1][i][j - 1]) % mod
tmp1 = 1
if i - 1 <= j and i > 0:
if i <= j - 1:
tmp1 += dp[1][i][j - 1]
tmp1 %= mod
tmp1 += dp[0][i - 1][j]
tmp1 %= mod
tmp2 = 1
if i <= j - 1 and j > 0:
if i - 1 <= j:
tmp2 += dp[1][i - 1][j]
tmp2 %= mod
tmp2 += dp[0][i][j - 1]
tmp2 %= mod
dp[1][i][j] = max(tmp1, tmp2) % mod
print(max(dp[0][-1][-1], dp[1][-1][-1]) % mod) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER VAR |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)
n = int(input())
mod = 10**9 + 7
ANS = [([0] * (n + 1)) for i in range(n + 1)]
def bra(x, y):
if ANS[x][y] != 0:
return ANS[x][y]
if x == y == 0:
ANS[x][y] = 0
return 0
if (x + y) % 2 == 1:
A = 1
else:
A = 0
if x == y:
ANS[x][y] = A + bra(x - 1, y)
return ANS[x][y]
elif x == 0:
ANS[x][y] = A + bra(x, y - 1)
return ANS[x][y]
elif y == 0:
ANS[x][y] = A + bra(x - 1, y)
return ANS[x][y]
elif x < y and x != 0 and y != 0:
ANS[x][y] = A + bra(x - 1, y) + bra(x, y - 1)
return ANS[x][y]
print(bra(n, n) % mod) | IMPORT ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_DEF IF VAR VAR VAR NUMBER RETURN VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR RETURN VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR RETURN VAR VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | n = int(input())
dp = [[(0) for i in range(2005)] for j in range(2005)]
ans, mod = 0, 10**9 + 7
dp[1][1] = 1
for i in range(1, 2 * n):
for j in range(2 * n):
dp[i + 1][j + 1] += dp[i][j]
if dp[i + 1][j + 1] >= mod:
dp[i + 1][j + 1] -= mod
if j > 0:
dp[i + 1][j - 1] += dp[i][j]
if dp[i + 1][j - 1] >= mod:
dp[i + 1][j - 1] -= mod
for i in range(1, 2 * n, 2):
for j in range(2 * n):
valid = i + j <= 2 * n
if valid:
ans += dp[i][j]
if ans >= mod:
ans -= mod
print(ans % mod) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR IF VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR IF VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP NUMBER VAR IF VAR VAR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | n = int(input())
board = [[(0) for i in range(n + 1)] for j in range(2 * n + 1)]
board[0][0] = 1
for i in range(1, n):
for j in range(len(board[0])):
if j > 0:
board[i][j - 1] += board[i - 1][j]
if j + 1 < len(board[0]) and j < 2 * n - i:
board[i][j + 1] += board[i - 1][j]
for i in range(n, 2 * n + 1):
for j in range(len(board[0])):
if j > 0:
board[i][j - 1] += board[i - 1][j]
if j + 1 < len(board[0]) and j < 2 * n - i:
board[i][j + 1] += board[i - 1][j]
ans = 0
for i in range(1, len(board), 2):
ans += sum(board[i])
ans %= 1000000007
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | import sys
sys.setrecursionlimit(10**9)
def fun(plus, cnt):
if check[plus][cnt]:
return check[plus][cnt]
val = 0
if plus < n and 2 * n - cnt > plus:
val += fun(plus + 1, cnt + 1)
if plus > 0:
val += fun(plus - 1, cnt + 1)
check[plus][cnt] = val
if cnt % 2:
check[plus][cnt] += 1
check[plus][cnt] %= 10**9 + 7
return check[plus][cnt]
n = int(input())
check = [[(0) for _ in range(2 * n + 1)] for _ in range(2 * n + 1)]
print(fun(1, 1) % (10**9 + 7)) | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER FUNC_DEF IF VAR VAR VAR RETURN VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR BIN_OP BIN_OP NUMBER VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR IF BIN_OP VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER RETURN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR NUMBER NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | from itertools import accumulate
N = int(input())
arr = [1]
res = 1
for i in range(2, N + 1):
arr = list(accumulate(arr))
arr = arr + [arr[-1]]
d = i + 1
s = 0
for i in range(len(arr)):
s += arr[i] * (d // 2)
d -= 1
res += s
res = res % 1000000007
print(res) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR LIST VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | n = int(input())
f = [([0] * (n + 1)) for i in range(n + 1)]
g = [([0] * (n + 1)) for i in range(n + 1)]
mod = 10**9 + 7
for i in range(1, n + 1):
f[0][i] = g[0][i - 1]
g[0][i] = f[0][i - 1] + 1
t = [0, 0]
for i in range(1, n + 1):
for j in range(i, n + 1):
if i > 0:
f[i][j] = g[i - 1][j]
t[0] = g[i - 1][j]
t[1] = f[i - 1][j] + 1
if j > i:
f[i][j] = (f[i][j] + g[i][j - 1]) % mod
t[0] = (t[0] + f[i][j - 1] + 1) % mod
t[1] = (t[1] + g[i][j - 1]) % mod
for k in t:
g[i][j] = max(g[i][j], k)
print(g[n][n] % mod) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR FOR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | import sys
u = 1000000007
def P(n):
X = [[(0) for i in range(2 * n + 1)] for j in range(2 * n + 1)]
X[0][0] = 1
r = 0
for i in range(1, 2 * n + 1):
X[i][0] = X[i - 1][1]
for j in range(1, min([2 * n + 1 - i, i + 1])):
X[i][j] = (X[i - 1][j + 1] + X[i - 1][j - 1]) % u
if i % 2:
r += sum(X[i])
return r % u
print(P(int(sys.stdin.read()[:-1]))) | IMPORT ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR LIST BIN_OP BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR RETURN BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER |
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...
It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.
The definition of correct bracket sequence is as follows: The empty sequence is a correct bracket sequence, If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence, If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence.
For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.
Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Input-----
The only line contains a single integer $n$ ($1 \le n \le 1000$).
-----Output-----
Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.
-----Examples-----
Input
1
Output
1
Input
2
Output
3
Input
3
Output
9
-----Note-----
The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image] | mod = 1000000007
n = int(input())
cat = [0] * 1100
f, c, i = 1, 1, 1
while i < 1100:
cat[i] = f
i += 1
c = c * (8 * i - 12) // i
f = c - f
cat = cat[1:-1]
sm = 0
for i in range(3, n + 3):
sm += (cat[i - 1] + (-1) ** (i - 1)) // (1 << i)
print(sm % mod) | ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER WHILE VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
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