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Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
t = input()
s = input()
zeros = 0
ones = 0
for i in s:
if i == "(":
zeros += 1
elif i == ")":
ones += 1
s = list(s)
i = 0
nbr_1 = 0
nbr_0 = 0
while i < len(s):
if nbr_1 >= nbr_0 and nbr_1 != 0:
break
if s[i] == "(":
nbr_0 += 1
zeros -= 1
elif s[i] == ")":
nbr_1 += 1
elif s[i] == "?" and nbr_0 + zeros < len(s) / 2:
s[i] = "("
nbr_0 += 1
else:
s[i] = ")"
nbr_1 += 1
i += 1
if nbr_0 != nbr_1 or nbr_0 < len(s) / 2:
print(":(")
else:
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
if n % 2 == 1:
print(":(")
exit(0)
s = list(input())
d = 0
count1 = s.count("?")
count2 = s.count("(")
count3 = s.count(")")
i = n - 1
while count2 - count3 > 0:
if i < 0:
print(":(")
exit(0)
if s[i] == "?":
s[i] = ")"
count3 += 1
count1 -= 1
i -= 1
i = 0
while count2 - count3 < 0:
if i == n:
print(":(")
exit(0)
if s[i] == "?":
s[i] = "("
count2 += 1
count1 -= 1
i += 1
p = count1 // 2
q = n - 1
d = 0
for i in range(n):
if s[i] == "(":
d += 1
elif s[i] == ")":
d -= 1
if d <= 0 and i < q:
print(":(")
exit(0)
elif p != 0:
s[i] = "("
d += 1
p -= 1
else:
s[i] = ")"
d -= 1
if d <= 0 and i < q:
print(":(")
exit(0)
if d > 0:
print(":(")
exit(0)
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR NUMBER WHILE BIN_OP VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
x = []
res = []
o1, c1 = 0, 0
for i in s:
x.append(i)
if i == "(":
o1 += 1
elif i == ")":
c1 += 1
if n % 2:
print(":(")
elif x[0] == ")" or x[n - 1] == "(":
print(":(")
else:
o, c = n / 2 - o1, n / 2 - c1
o2, c2 = 0, 0
for i in range(n):
if x[i] == "?":
if o > 0:
res.append("(")
o -= 1
o1 += 1
o2 += 1
else:
res.append(")")
c -= 1
c1 += 1
c2 += 1
else:
res.append(x[i])
if x[i] == "(":
o2 += 1
else:
c2 += 1
if i != n - 1 and c2 == o2:
break
if c2 == o2 and c2 + o2 == n:
for i in range(n):
print(res[i], end="")
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR IF VAR VAR BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
open_list = ["("]
close_list = [")"]
def check(myStr):
stack = []
for i in myStr:
if i in open_list:
stack.append(i)
elif i in close_list:
pos = close_list.index(i)
if len(stack) > 0 and open_list[pos] == stack[len(stack) - 1]:
stack.pop()
else:
return 0
if len(stack) == 0:
return 1
n = int(input())
s = input()
left = s.count("(")
right = s.count(")")
re_left = n // 2 - left
re_right = n // 2 - right
arr = list(s)
for i in range(len(arr)):
if arr[i] == "?":
if re_left > 0:
arr[i] = "("
re_left -= 1
elif re_right > 0:
arr[i] = ")"
re_right -= 1
ct_le = 0
ct_ri = 0
flag = 0
for i in range(len(arr)):
if arr[i] == "(":
ct_le += 1
else:
ct_ri += 1
if ct_le == ct_ri and i != len(arr) - 1:
flag = 1
s1 = "".join(arr)
if flag == 0 and ct_le == ct_ri and check(s1):
print("".join(arr))
else:
print(":(")
|
ASSIGN VAR LIST STRING ASSIGN VAR LIST STRING FUNC_DEF ASSIGN VAR LIST FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR RETURN NUMBER IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR IF VAR NUMBER VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input().strip()
if n % 2 != 0:
print(":(")
exit()
upper = 0
uppers = []
lower = 0
for c in s[:-1]:
if c == "(":
upper += 1
lower += 1
elif c == ")":
upper -= 1
lower -= 1
if upper <= 0:
print(":(")
exit()
else:
upper += 1
lower -= 1
if lower == 0:
lower = 2
if lower == -1:
lower = 1
uppers.append(upper)
if lower != 1:
print(":(")
exit()
if s[-1] == "(":
print(":(")
exit()
arr = [")"]
height = 1
for i in range(n - 2, 0, -1):
if s[i] == "(":
arr.append("(")
height -= 1
elif s[i] == ")":
arr.append(")")
height += 1
elif uppers[i - 1] >= height + 1:
arr.append(")")
height += 1
else:
arr.append("(")
height -= 1
print("(" + "".join(reversed(arr)))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR LIST STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR BIN_OP STRING FUNC_CALL STRING FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
kk = lambda: map(int, input().split())
ll = lambda: list(kk())
n, s = int(input()), list(input())
if s[-1] == "(" or s[0] == ")" or n % 2 == 1:
print(":(")
exit()
qcount = ocount = ccount = 0
for i in range(n):
if s[i] == "?":
qcount += 1
elif s[i] == "(":
ocount += 1
else:
ccount += 1
res = []
if s[0] == "(":
ocount -= 1
else:
qcount -= 1
if s[-1] == ")":
ccount -= 1
else:
qcount -= 1
val = 0
for i in range(1, n - 1):
if val < 0:
print(":(")
exit()
if s[i] == ")":
res.append(")")
val -= 1
ccount -= 1
elif s[i] == "(":
val += 1
res.append("(")
ocount -= 1
else:
if val + ocount >= ccount + qcount:
res.append(")")
val -= 1
else:
res.append("(")
val += 1
qcount -= 1
if val != 0:
print(":(")
exit()
print("(" + "".join(res) + ")")
|
ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER STRING VAR NUMBER STRING BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR LIST IF VAR NUMBER STRING VAR NUMBER VAR NUMBER IF VAR NUMBER STRING VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP STRING FUNC_CALL STRING VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
l = list(input())
if n % 2 == 1:
print(":(")
exit()
if n == 2:
if l[0] == ")" or l[1] == "(":
print(":(")
else:
print("()")
exit()
if l[0] == ")" or l[1] == ")" or l[-1] == "(" or l[-2] == "(":
print(":(")
exit()
l[0] = "("
l[1] = "("
l[-1] = ")"
l[-2] = ")"
p = l.count(")") - l.count("(")
q = l.count("?")
a = (q + p) // 2
b = q - a
cnt = 0
for i in range(len(l)):
if l[i] == "?":
if cnt < a:
l[i] = "("
else:
l[i] = ")"
cnt += 1
cnt = 0
for p, i in enumerate(l):
if i == "(":
cnt += 1
else:
cnt -= 1
if cnt == 0 and p != n - 1:
print(":(")
exit()
else:
if cnt != 0:
print(":(")
exit()
print("".join(l))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER STRING VAR NUMBER STRING VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR STRING FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR VAR ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
N = int(input())
S = input()
left = S.count("(")
right = S.count(")")
plus = 0
ans = ""
for i in range(N):
if S[i] == "?":
if left < N // 2:
ans += "("
plus += 1
left += 1
else:
ans += ")"
plus -= 1
right += 1
elif S[i] == "(":
ans += "("
plus += 1
else:
ans += ")"
plus -= 1
if i != N - 1 and plus <= 0:
print(":(")
exit()
if left == right == N // 2:
print(ans)
else:
print(":(")
exit()
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER VAR STRING VAR NUMBER VAR NUMBER VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR STRING VAR NUMBER VAR STRING VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
if n % 2:
print(":(")
exit(0)
s = list(input())
ans = ""
c1, c2 = s.count("("), s.count(")")
cnt90 = 0
cnt9, cnt0 = 0, 0
for i in s[:-1]:
if i == "(":
cnt90 += 1
if cnt9 + cnt90 > n // 2:
print(":(")
exit(0)
ans += i
elif i == ")":
cnt0 += 1
if cnt0 > n // 2:
print(":(")
exit(0)
ans += i
elif cnt9 + c1 < n // 2:
ans += "("
cnt9 += 1
else:
ans += ")"
cnt0 += 1
if cnt90 + cnt9 <= cnt0:
print(":(")
exit(0)
if s[-1] == "(" or cnt9 + cnt90 != n // 2:
print(":(")
exit(0)
print(ans + ")")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR VAR FUNC_CALL VAR STRING FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR NUMBER IF VAR STRING VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER VAR VAR IF VAR STRING VAR NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER VAR VAR IF BIN_OP VAR VAR BIN_OP VAR NUMBER VAR STRING VAR NUMBER VAR STRING VAR NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER STRING BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def solve():
n = int(input())
s = list(input())
if n % 2:
print(":(")
return
cv = s.count("?")
co = s.count("(")
cc = s.count(")")
b = 0
for i in range(n):
if s[i] == "(":
b += 1
elif s[i] == ")":
b -= 1
elif co < n // 2:
s[i] = "("
b += 1
co += 1
else:
s[i] = ")"
b -= 1
cc += 1
if b <= 0 and i != n - 1:
print(":(")
return
if b != 0:
print(":(")
return
print("".join(s))
solve()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN IF VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
o = 0
work = True
s = list(input())
a = [0] * n
op = [0] * n
o = 0
c = 0
k = n - 1
for i in s[::-1]:
a[k] = c
op[k] = o
k -= 1
if i != "(":
c += 1
else:
o += 1
o = 0
for j, i in enumerate(s):
if o == 0 and j != 0:
work = False
break
if i == "(":
o += 1
elif i == ")":
o -= 1
if o <= 0 and j != n - 1:
work = False
break
elif a[j] > o + op[j]:
s[j] = "("
o += 1
else:
s[j] = ")"
o -= 1
if work and o == 0:
print("".join(s))
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
if n % 2:
print(":(")
exit()
ans = ""
o = 0
c = 0
for i in s:
if i == "(":
o += 1
elif i == ")":
c += 1
if o > n // 2 or c > n // 2:
print(":(")
exit()
o = n // 2 - o
for i in s:
if i == "?":
if o:
o -= 1
ans += "("
else:
ans += ")"
else:
ans += i
for i in range(n - 1):
if ans[i] == "(":
o += 1
else:
o -= 1
if o <= 0:
print(":(")
exit()
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR FOR VAR VAR IF VAR STRING IF VAR VAR NUMBER VAR STRING VAR STRING VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
import sys
n = int(input())
s = input()
if n % 2 == 1:
print(":(")
sys.exit(0)
d = {}
d["("] = 1
d[")"] = -1
d["?"] = 0
suma = 0
cale = 0
dop = [[0, 0] for i in range(n)]
dop[n - 1] = [0, 0]
for i in range(1, n):
j = n - i - 1
if s[j + 1] == ")":
dop[j] = [max(1, dop[j + 1][0] + 1), dop[j + 1][1] + 1]
if s[j + 1] == "(":
dop[j] = [max(1, dop[j + 1][0] - 1), dop[j + 1][1] - 1]
if s[j + 1] == "?":
dop[j] = [max(1, dop[j + 1][0] - 1), dop[j + 1][1] + 1]
if dop[j][0] > dop[j][1]:
print(":(")
sys.exit(0)
suma = 0
odp = []
pyk = 0
for i in range(n):
if s[i] != "?":
odp.append(s[i])
if s[i] == "(":
suma += 1
if s[i] == ")":
pyk += 1
suma -= 1
elif suma + 1 >= dop[i][0] and suma + 1 <= dop[i][1]:
odp.append("(")
suma += 1
elif suma - 1 >= dop[i][0] and suma - 1 <= dop[i][1]:
odp.append(")")
suma -= 1
else:
print(":(")
sys.exit(0)
if pyk > n // 2:
print(":(")
else:
for i in range(n):
print(odp[i], end="")
|
IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR DICT ASSIGN VAR STRING NUMBER ASSIGN VAR STRING NUMBER ASSIGN VAR STRING NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR LIST FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR LIST FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR LIST FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
l = int(input())
s = list(input())
count = s.count("(")
pair_count = 0
if l % 2 != 0:
print(":(")
exit(0)
if count > l // 2:
print(":(")
exit(0)
for ind in range(l):
if s[ind] == "?":
if count < l // 2:
s[ind] = "("
count += 1
else:
s[ind] = ")"
if s[ind] == "(":
pair_count += 1
if s[ind] == ")":
pair_count -= 1
if pair_count <= 0 and ind + 1 != l:
print(":(")
exit(0)
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
if n % 2:
print(":(")
quit()
l = n // 2 - s.count("(")
for i in range(n):
if s[i] == "?":
if l > 0:
s[i] = "("
l -= 1
else:
s[i] = ")"
bal = 0
if s[0] != "(" or s[-1] != ")":
print(":(")
quit()
for i in range(n):
if s[i] == "(":
bal += 1
if s[i] == ")":
bal -= 1
if bal == 0 and i != n - 1:
print(":(")
quit()
if bal != 0 and i == n - 1:
print(":(")
quit()
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
if n % 2:
print(":(")
else:
k = ""
b = [0] * n
open1 = 0
close1 = 0
ind = n
mn = 10**9
open2 = 0
close2 = 0
for i in range(n):
if s[i] == ")":
close1 += 1
elif s[i] == "(":
open1 += 1
for i in range(n):
if s[i] == "?":
if open1 < n // 2:
k += "("
open1 += 1
open2 += 1
else:
k += ")"
close1 += 1
close2 += 1
else:
k += s[i]
if k[i] == ")":
close2 += 1
else:
open2 += 1
b[i] = open2 - close2
if b[i] == 0:
ind = min(ind, i)
mn = min(b[i], mn)
if ind == n - 1 and mn == 0:
print(k)
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR STRING ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER VAR STRING VAR NUMBER VAR NUMBER VAR STRING VAR NUMBER VAR NUMBER VAR VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR IF VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
if n % 2 != 0:
print(":(")
else:
total_left = n / 2 - s.count("(")
k = []
c, i = 0, 0
while i < n:
if s[i] == "(":
k.append("(")
c += 1
elif s[i] == ")":
k.append(")")
c -= 1
elif total_left > 0:
total_left -= 1
k.append("(")
c += 1
else:
k.append(")")
c -= 1
if c < 0:
print(":(")
break
if c == 0 and i != n - 1:
print(":(")
break
elif i == n - 1 and c != 0:
print(":(")
break
elif i == n - 1 and c == 0:
print("".join(k))
i += 1
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR VAR NUMBER NUMBER WHILE VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR NUMBER
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
if n % 2:
print(":(")
return
s = [c for c in input()]
if s[0] == ")" or s[-1] == "(":
print(":(")
return
s[0] = "("
s[-1] = ")"
count_open = s.count("(")
count_close = s.count(")")
half = n // 2
if count_close > half or count_open > half:
print(":(")
return
available = half - count_open
current = 0
for i, x in enumerate(s):
if x == "(":
current += 1
elif x == ")":
current -= 1
elif available:
current += 1
available -= 1
s[i] = "("
else:
current -= 1
s[i] = ")"
if current == 0 and i != n - 1:
print(":(")
return
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR STRING RETURN ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING RETURN ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER IF VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
a = list(input())
if n % 2:
print(":(")
exit(0)
cnt = 0
osc = a.count("(")
ans = osc <= n // 2
for i in range(len(a)):
if a[i] == "?":
if osc == n // 2:
a[i] = ")"
cnt -= 1
else:
a[i] = "("
cnt += 1
osc += 1
elif a[i] == "(":
cnt += 1
else:
cnt -= 1
if cnt <= 0 and i != len(a) - 1:
ans = False
if ans:
for i in a:
print(i, end="")
elif not ans or cnt != 0:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING IF VAR VAR NUMBER EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
import sys
length = int(input())
op = 0
ret = "("
word = input()
if length % 2 == 1 or word[0] == ")" or word[-1] == "(":
print(":(")
sys.exit()
word2 = word[1:-1]
mc = word2.count(")")
mo = word2.count("(")
goal = (length - 2) // 2
nc = goal - mc
no = goal - mo
if nc < 0 or no < 0:
print(":(")
sys.exit()
for l in word2:
if l == "(":
ret += "("
op += 1
elif l == ")":
ret += ")"
op -= 1
elif no > 0:
no -= 1
ret += "("
op += 1
else:
nc -= 1
ret += ")"
op -= 1
if op == -1:
print(":(")
sys.exit()
if op == 0:
ret += ")"
print(ret)
else:
print(":(")
|
IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR VAR IF VAR STRING VAR STRING VAR NUMBER IF VAR STRING VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER VAR STRING VAR NUMBER VAR NUMBER VAR STRING VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER VAR STRING EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
sctype = {"(": 0, ")": 1, "?": -1}
ln = int(input())
s = list(input())
if s[0] == ")" or ln % 2 == 1:
print(":(")
exit()
s[0] = "("
opp = 0
opl = 1
op = []
r = True
pspt = [0, 0]
for i in range(ln):
if sctype[s[i]] == -1:
opl += 1
else:
pspt[sctype[s[i]]] += 1
op.append([opp, opl, list(pspt)])
opp = i
opl = 0
op.append([ln, opl, [max(pspt), max(pspt)]])
nado = ln - max(pspt) * 2
nop = max(0, pspt[1] - pspt[0]) + nado // 2
ncl = max(0, pspt[0] - pspt[1]) + nado // 2
opp = -1
pspt = [0, 0]
for i in range(ln):
if sctype[s[i]] == -1:
if nop > 0:
nop -= 1
pspt[0] += 1
s[i] = "("
elif ncl > 0:
ncl -= 1
s[i] = ")"
pspt[1] += 1
else:
r = False
else:
opp += 1
pspt[sctype[s[i]]] += 1
if pspt[0] == pspt[1] and i + 1 != ln:
r = False
break
if r and pspt[0] == pspt[1]:
print(*s, sep="")
else:
print(":(")
|
ASSIGN VAR DICT STRING STRING STRING NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER STRING BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR LIST VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR LIST VAR VAR LIST FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER NUMBER ASSIGN VAR VAR STRING IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER IF VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
kol = int(input())
s = input()
leftB = s.count("(")
rightB = s.count(")")
q = s.count("?")
sres = ""
for i in range(kol):
if s[i] == "?":
if leftB * 2 < kol:
sres += "("
leftB += 1
else:
sres += ")"
else:
sres += s[i]
count = 0
for i in range(kol):
count += 1 if sres[i] == "(" else -1
if count < 1 and i < kol - 1:
print(":(")
exit()
elif count != 0 and i == kol - 1:
print(":(")
exit()
print(sres)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF BIN_OP VAR NUMBER VAR VAR STRING VAR NUMBER VAR STRING VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR STRING NUMBER NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input(""))
s = list(input(""))
x = 0
y = 0
p = 0
q = 0
z = 0
m = 0
for i in range(n):
if s[i] == "(":
m += 1
if n % 2 == 1:
print(":(")
else:
for i in range(n - 1):
if s[-i - 1] == "(":
x += 1
else:
y += 1
if s[i] == ")":
p += 1
else:
q += 1
if x >= y:
z = 1
break
if p >= q:
z = 1
break
if z == 1:
print(":(")
else:
for i in range(n):
if s[i] == "?":
if m >= n // 2:
s[i] = ")"
else:
s[i] = "("
m += 1
print(s[i], end="")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def mp():
return map(int, input().split())
n = int(input())
s = list(input())
b = 0
o = n // 2 - s.count("(")
for i in range(n):
if s[i] == "(":
b += 1
elif s[i] == ")":
b -= 1
elif o > 0:
s[i] = "("
b += 1
o -= 1
else:
s[i] = ")"
b -= 1
b = 0
fail = False
for i in range(n):
if s[i] == "(":
b += 1
else:
b -= 1
if b < 0 or i != n - 1 and b == 0:
fail = True
if fail or b != 0:
print(":(")
else:
print("".join(s))
|
FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def solve(s, n):
if n % 2 == 1:
return None
a = n // 2 - s.count("(")
b = n // 2 - s.count(")")
if a < 0 or b < 0:
return None
x = []
for e in s:
if e == "?":
if a > 0:
x.append("(")
a -= 1
else:
x.append(")")
b -= 1
else:
x.append(e)
balance = 0
for i, e in enumerate(x):
balance += 1 if e == "(" else -1
if balance <= 0 and i < n - 1:
return None
return "".join(x)
n = int(input())
s = input()
res = solve(s, n)
if res:
print(res)
else:
print(":(")
|
FUNC_DEF IF BIN_OP VAR NUMBER NUMBER RETURN NONE ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER RETURN NONE ASSIGN VAR LIST FOR VAR VAR IF VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR VAR STRING NUMBER NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER RETURN NONE RETURN FUNC_CALL STRING VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
_str = str
n, str = int(input()), list(input())
if str[0] == "?":
str[0] = "("
if str[-1] == "?":
str[-1] = ")"
pp, nn = n // 2, n // 2
for ch in str:
if ch == "(":
pp -= 1
if ch == ")":
nn -= 1
if n % 2 == 1 or str[0] != "(" or str[-1] != ")" or pp < 0 or nn < 0:
print(":(")
exit(0)
res, sum = [], 0
for ch in str:
if ch == "(":
res.append("(")
sum += 1
elif ch == ")":
res.append(")")
sum -= 1
elif pp > 0:
res.append("(")
sum += 1
pp -= 1
elif nn > 0:
res.append(")")
sum -= 1
nn -= 1
else:
print(":(")
exit(0)
if sum <= 0:
break
if sum == 0 and len(res) == n:
print(
_str(res).replace(", ", "").replace("[", "").replace("]", "").replace("'", "")
)
else:
print(":(")
|
ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER STRING ASSIGN VAR NUMBER STRING IF VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING VAR NUMBER STRING VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR LIST NUMBER FOR VAR VAR IF VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER IF VAR NUMBER FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL FUNC_CALL FUNC_CALL FUNC_CALL VAR VAR STRING STRING STRING STRING STRING STRING STRING STRING EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def main():
n = int(input())
s = input()
n_open = s.count("(")
n_close = s.count(")")
n_blank = s.count("?")
need_n_open = len(s) / 2 - n_open
if need_n_open < 0:
print(":(")
return
res = ""
cnt = 0
for i in range(len(s)):
c = s[i]
if c == "(":
cnt += 1
res += c
elif c == ")":
cnt -= 1
res += c
elif c == "?":
if need_n_open > 0:
res += "("
cnt += 1
need_n_open -= 1
else:
res += ")"
cnt -= 1
if cnt < 0:
break
if cnt == 0 and i < n - 1:
res = ":("
break
if cnt != 0:
res = ":("
print(res)
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR STRING VAR NUMBER VAR VAR IF VAR STRING VAR NUMBER VAR VAR IF VAR STRING IF VAR NUMBER VAR STRING VAR NUMBER VAR NUMBER VAR STRING VAR NUMBER IF VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR STRING IF VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
s = list(s)
a = n // 2 - s.count("(")
b = n - a - s.count(")")
for i in range(n):
if s[i] == "?":
s[i] = "("
a -= 1
if a <= 0:
break
for i in range(n):
if s[i] == "?":
s[i] = ")"
b -= 1
if b <= 0:
break
c = 0
for i in range(n):
if s[i] == "(":
c += 1
else:
c -= 1
if c <= 0 and i < n - 1:
c = 1
break
if c != 0:
print(":(")
else:
for i in s:
print(i, end="")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
l = s.count("(")
r = s.count(")")
for i in range(n):
if s[i] == "?":
if 2 * l < n:
l += 1
s[i] = "("
else:
s[i] = ")"
cnt = 0
for i in range(n):
cnt += 1 if s[i] == "(" else -1
if i < n - 1 and cnt < 1:
print(":(")
exit()
if i == n - 1 and cnt != 0:
print(":(")
exit()
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR STRING NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
st = list(input())
if n % 2 == 1:
print(":(")
else:
open = st.count("(")
close = 0
if open > n // 2:
print(":(")
else:
for e in range(len(st)):
if st[e] == "?":
if open < n / 2:
st[e] = "("
open += 1
else:
st[e] = ")"
op = 0
cl = 0
for e in range(len(st)):
if st[e] == "(":
op += 1
elif st[e] == ")":
cl += 1
if cl >= op and e != 0 and e != len(st) - 1:
print(":(")
exit(0)
print("".join(st))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
l, r = s.count("("), s.count(")")
for i in range(n):
if s[i] == "?" and 2 * l < n:
s[i] = "("
l += 1
elif s[i] == "?" and 2 * l >= n:
s[i] = ")"
cnt = 0
for i in range(n):
cnt = cnt + 1 if s[i] == "(" else cnt - 1
if cnt <= 0 and i < n - 1 or cnt > 0 and i == n - 1:
print(":(")
break
else:
print("".join(s) if n % 2 == 0 else ":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR STRING FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING BIN_OP NUMBER VAR VAR ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR STRING BIN_OP NUMBER VAR VAR ASSIGN VAR VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR STRING BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL STRING VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
if n % 2 == 1 or s[0] == ")" or s[len(s) - 1] == "(":
print(":(")
else:
q = s.count("?")
dl = s.count("(")
dr = s.count(")")
t = (dl + dr + q) // 2
if dl > dr:
t = t - dl
dl = t
dr = q - t
elif dr > dl:
t = t - dr
dr = t
dl = q - t
else:
dr = q // 2
dl = q // 2
for i in range(len(s)):
if s[i] == "?":
if dl > 0:
s[i] = "("
dl -= 1
elif dr > 0:
s[i] = ")"
dr -= 1
if dr == dl and dl == 0:
break
st = [s[0]]
fail = False
for i in range(1, len(s)):
if st == []:
fail = True
break
elif s[i] == "(":
st.append(i)
elif s[i] == ")":
st.pop()
if fail or len(st) > 0:
print(":(")
else:
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING VAR BIN_OP FUNC_CALL VAR VAR NUMBER STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR LIST VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR LIST ASSIGN VAR NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR IF VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def readint():
return int(input())
n = readint()
s = input()
a = 0
b = 0
for c in s:
if c == "(":
a += 1
elif c == ")":
b += 1
x = n - (a + b)
m = a - b
p = x - m
p /= 2
r = []
for c in s:
if c == "?":
if p:
p -= 1
r += "("
else:
r += ")"
else:
r.append(c)
m = 0
ok = True
for i in range(n):
if r[i] == "(":
m += 1
else:
m -= 1
if m < 0 or m == 0 and i != n - 1:
ok = False
if m != 0:
ok = False
if ok:
print("".join(r))
else:
print(":(")
|
FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR LIST FOR VAR VAR IF VAR STRING IF VAR VAR NUMBER VAR STRING VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
state = list(input())
if ")" == state[0] or "(" == state[-1] or len(state) % 2 == 1:
print(":(")
else:
state[0] = "("
state[-1] = ")"
left_index = 1
right_index = 1
question_index = 1
while (
left_index != len(state) - 1
and right_index != len(state) - 1
and question_index != len(state) - 1
):
while left_index < len(state) - 1:
if state[left_index] == ")":
break
left_index += 1
if left_index == len(state) - 1:
break
while right_index < left_index:
if state[right_index] == "(":
break
right_index += 1
if right_index == left_index:
while question_index < right_index:
if state[question_index] == "?":
break
question_index += 1
if question_index == right_index:
print(":(")
exit()
else:
state[question_index] = "("
left_index += 1
right_index += 1
question_index += 1
left_index = len(state) - 2
right_index = len(state) - 2
question_index = len(state) - 2
while left_index != 0 and right_index != 0 and question_index != 0:
while right_index > 0:
if state[right_index] == "(":
break
right_index -= 1
if right_index == 0:
break
while right_index < left_index:
if state[left_index] == ")":
break
left_index -= 1
if right_index == left_index:
while question_index > right_index:
if state[question_index] == "?":
break
question_index -= 1
if question_index == right_index:
print(":(")
exit()
else:
state[question_index] = ")"
left_index -= 1
right_index -= 1
question_index -= 1
next_ele = "("
for i in range(len(state)):
if state[i] == "?":
state[i] = next_ele
next_ele = ")" if next_ele == "(" else "("
print("".join(state))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF STRING VAR NUMBER STRING VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR WHILE VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR NUMBER VAR NUMBER VAR NUMBER WHILE VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR WHILE VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR ASSIGN VAR VAR STRING STRING STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = [i for i in input()]
o = 0
z = 0
O = s.count("(")
Z = s.count(")")
V = s.count("?")
on = (V - (O - Z)) // 2
flag = 0
if on >= 0:
for i in range(n):
if s[i] == "(":
o += 1
if s[i] == ")":
z += 1
if s[i] == "?":
if on > 0:
s[i] = "("
o += 1
on -= 1
else:
s[i] = ")"
z += 1
if z == o and i != len(s) - 1:
flag = 1
break
if z > o:
flag = 1
break
else:
flag = 1
if z != o:
flag = 1
if flag == 0:
print(*s, sep="")
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
b = s.count("(") - s.count(")")
undef = s.count("?")
if b != 0 and undef == 0 or b > undef:
print(":(")
exit()
if b >= 0:
nClose = b
undef -= nClose
if undef & 1 == 1:
print(":(")
exit()
nOpen = undef >> 1
nClose += nOpen
else:
if undef + b < 0:
print(":(")
exit()
nOpen = -b
undef -= nOpen
if undef & 1 == 1:
print(":(")
exit()
nClose = undef >> 1
nOpen += nClose
b = 0
for i in range(n):
if s[i] == "(":
b += 1
elif s[i] == ")":
b -= 1
elif nOpen > 0:
s[i] = "("
nOpen -= 1
b += 1
else:
s[i] = ")"
nClose -= 1
b -= 1
if i < n - 1 and b <= 0:
print(":(")
exit()
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR STRING FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER ASSIGN VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
i = 0
j = n - 1
flag = 0
if s[0] == ")" or n % 2 != 0 or s[-1] == "(":
flag = 1
else:
z = n // 2 - s.count("(")
for i in range(n):
if s[i] == "?":
if z != 0:
s[i] = "("
z -= 1
else:
s[i] = ")"
for i in range(n):
if s[i] == "(":
s[i] = 1
else:
s[i] = -1
k = sum(s)
if k != 0:
flag = 1
else:
sum = 0
for i in range(n - 1):
sum += s[i]
if sum == 0:
flag = 1
break
if flag == 1:
print(":(")
else:
for i in range(n):
if s[i] == 1:
s[i] = "("
else:
s[i] = ")"
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER STRING BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def main():
n = int(input())
s = list(input())
temp_count = 0
for c in s:
if c == "(":
temp_count += 1
stack = []
for i in range(n):
if i == n - 1 and (len(stack) != 1 or s[i] == "("):
print(":(")
return
if s[i] == "(":
stack.append(s[i])
elif s[i] == ")":
if not stack:
print(":(")
return
stack.pop()
elif i != n - 1 and temp_count < n // 2:
s[i] = "("
stack.append("(")
temp_count += 1
else:
s[i] = ")"
stack.pop()
if i != n - 1 and not stack:
print(":(")
return
print("".join(s))
def __starting_point():
main()
__starting_point()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR STRING EXPR FUNC_CALL VAR STRING RETURN IF VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR IF VAR VAR STRING IF VAR EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR FUNC_CALL STRING VAR FUNC_DEF EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
res = [""] * n
if n % 2 == 1 or s[0] == ")" or s[n - 1] == "(":
print(":(")
else:
count = 0
for i, c in enumerate(s):
res[i] = c
if c == ")":
count += 1
for i in range(n - 1, -1, -1):
if s[i] == "?":
if count < n / 2:
res[i] = ")"
count += 1
else:
res[i] = "("
count = 0
for i, el in enumerate(res):
if count < 0 or count == 0 and i != len(res) - 1 and i != 0:
res = []
break
if el == "(":
count += 1
if el == ")":
count -= 1
if count == 0 and res:
print("".join(res))
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST STRING VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR LIST IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER IF VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
A = list(input())
if A[0] == ")" or A[n - 1] == "(":
print(":(")
else:
A[0] = "("
A[n - 1] = ")"
a = A.count("(")
b = 0
c = 0
for i in range(n):
if A[i] == "(":
b += 1
elif A[i] == ")":
b -= 1
elif c < n // 2 - a:
A[i] = "("
c += 1
b += 1
else:
A[i] = ")"
b -= 1
if b == 0 and i != n - 1:
b = 1
break
if b != 0:
print(":(")
else:
print("".join(A))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER STRING ASSIGN VAR BIN_OP VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
a = int(input())
b = input()
if a > 2 and a % 2 == 0 and b[0] != ")" and b[-1] != "(":
c = 0
d = 0
e = True
k = ""
i = 1
q = 0
while i < a - 1 and e == True:
if b[i] == "(":
c += 1
elif b[i] == ")":
c -= 1
else:
d += 1
if c == 0:
if d != 0:
d -= 1
q += 1
c += 1
elif c < 0:
if d > 0:
d -= 1
q += 1
c += 1
if d > 1:
d -= 1
q += 1
c += 1
else:
e = False
i += 1
t = c
u = b[1 : a - 1].count("?")
u = u - t - q
if u >= 0 and e == True:
k += "("
w = 1
for i in range(1, a - 1):
if b[i] == "(":
k += "("
elif b[i] == ")":
k += ")"
elif q > 0:
k += "("
q -= 1
elif u > 0:
if w == 1:
k += "("
else:
k += ")"
w *= -1
u -= 1
else:
k += ")"
k += ")"
c = 0
i = 1
while i < a - 1 and e == True:
if k[i] == "(":
c += 1
else:
c -= 1
if c < 0:
e = False
i += 1
if e == False:
print(":(")
else:
print(k)
else:
print(":(")
elif a == 2:
if b[0] != ")" and b[1] != "(":
print("()")
else:
print(":(")
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING VAR NUMBER STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR STRING VAR STRING IF VAR VAR STRING VAR STRING IF VAR NUMBER VAR STRING VAR NUMBER IF VAR NUMBER IF VAR NUMBER VAR STRING VAR STRING VAR NUMBER VAR NUMBER VAR STRING VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING IF VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
if n % 2 == 1 or s[0] not in ["(", "?"]:
print(":(")
else:
l, r = s.count("("), s.count(")")
if l > n // 2 or r > n // 2:
print(":(")
else:
pl, pr = abs(n // 2 - l), abs(n // 2 - r)
cnt = 1
sc = "("
p = 1 if s[0] == "?" else 0
i = 1
while i < n:
if s[i] == "(":
cnt += 1
sc += "("
elif s[i] == ")":
cnt -= 1
sc += ")"
elif s[i] == "?" and p < pl:
cnt += 1
p += 1
sc += "("
elif s[i] == "?" and p >= pl:
cnt -= 1
sc += ")"
if cnt == 0 and i < n - 1:
print(":(")
break
i += 1
else:
print(sc)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER VAR NUMBER LIST STRING STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR STRING FUNC_CALL VAR STRING IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR VAR NUMBER STRING NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR NUMBER VAR STRING IF VAR VAR STRING VAR NUMBER VAR STRING IF VAR VAR STRING VAR VAR VAR NUMBER VAR NUMBER VAR STRING IF VAR VAR STRING VAR VAR VAR NUMBER VAR STRING IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
from sys import stdin, stdout
N = int(input())
s = input()
if N % 2 == 1:
print(":(")
quit()
level = 0
left = 0
for i in range(N):
if s[i] == "(":
left += 1
s = list(s)
for i in range(N):
if s[i] == "(":
level += 1
elif s[i] == ")":
level -= 1
elif s[i] == "?":
if left < N // 2:
s[i] = "("
left += 1
level += 1
else:
s[i] = ")"
level -= 1
if level <= 0 and i != N - 1:
print(":(")
quit()
if level != 0:
print(":(")
quit()
else:
print(*s, sep="")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = q = int(input())
k = list(input())
cntl = k.count("(")
cntr = k.count(")")
cntq = k.count("?")
for i in range(n):
if k[i] == "?":
if cntl < q // 2 and cntr + cntq >= q // 2:
k[i] = "("
cntl += 1
cntq -= 1
else:
k[i] = ")"
cntr += 1
cntq -= 1
def check():
cnt = 0
m = 0
for i in k:
m += 1
if i == "(":
cnt += 1
else:
cnt -= 1
if cnt == 0 and m < n or cnt < 0:
return False
return cnt == 0
print("".join(k) if check() else ":(")
|
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER RETURN VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
import sys
input = sys.stdin.readline
n = int(input())
s = list(input().rstrip())
if n % 2:
print(":(")
exit()
cnt = 0
c = 0
for i in range(n):
if s[i] == "?":
c += 1
elif s[i] == "(":
cnt += 1
a = n // 2 - cnt
b = n // 2 - (n - cnt - c)
if a < 0 or b < 0:
print(":(")
exit()
for i in range(n):
if s[i] == "?":
if a:
s[i] = "("
a -= 1
else:
s[i] = ")"
cnt = 0
f = True
for i in range(n):
if s[i] == "(":
cnt += 1
else:
cnt -= 1
if cnt < 0:
f = False
if cnt == 0 and f and i < n - 1:
print(":(")
exit()
elif cnt == 0 and f:
print("".join(s))
exit()
print(":(")
|
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
S = list(input())
def solve():
if n % 2:
return ":("
L = [0] * n
maxcnt = n // 2 - S.count("(")
cnt = 0
for i in range(n):
if S[i] == "(":
L[i] = 1
elif S[i] == ")":
L[i] = -1
elif cnt < maxcnt:
L[i] = 1
S[i] = "("
cnt += 1
else:
L[i] = -1
S[i] = ")"
judge = 0
for i in range(n - 1):
judge += L[i]
if judge <= 0:
return ":("
judge += L[-1]
if judge != 0:
return ":("
return "".join(S)
print(solve())
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF BIN_OP VAR NUMBER RETURN STRING ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR NUMBER RETURN STRING VAR VAR NUMBER IF VAR NUMBER RETURN STRING RETURN FUNC_CALL STRING VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def fix(S):
N = len(S)
s = 0
minS = 0
for n in range(N):
if S[n] == "(":
s += 1
elif S[n] == ")":
s -= 1
minS = min(minS, s)
if minS < 0:
for n in range(N):
if S[n] == "?":
S[n] = "("
minS += 1
s += 1
if not minS:
break
if s > 0:
for n in range(N - 1, -1, -1):
if S[n] == "?":
S[n] = ")"
s -= 1
if not s:
break
nq = 0
for n in range(N):
if S[n] == "?":
nq += 1
if nq % 2:
return False
added = 0
for n in range(N):
if S[n] == "?":
if added < nq // 2:
S[n] = "("
else:
S[n] = ")"
added += 1
s = 0
for n in range(N):
if S[n] == "(":
s += 1
else:
s -= 1
if s < 0:
return False
if s:
return False
return S
def sv():
N = int(input())
S = list(input())
if N % 2:
return False
if S[0] == ")" or S[N - 1] == "(":
return False
if N == 2:
print("()")
return True
f = fix(S[1:-1])
if not f:
return False
print("(%s)" % "".join(f))
return True
if not sv():
print(":(")
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR IF VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER IF VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF BIN_OP VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER RETURN NUMBER IF VAR RETURN NUMBER RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER RETURN NUMBER IF VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING RETURN NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR RETURN NUMBER EXPR FUNC_CALL VAR BIN_OP STRING FUNC_CALL STRING VAR RETURN NUMBER IF FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = str(input()[:n])
a = [0] * n
ocount = 0
ccount = 0
notfulfilled = 0
pos = 1
if n % 2 != 0:
print(":(")
else:
for i in range(n):
if s[i] == "(":
a[i] = 1
ocount += 1
elif s[i] == ")":
a[i] = -1
ccount += 1
for pos in range(n - 1):
if a[pos] == 1:
notfulfilled += 1
elif a[pos] == 0:
if ocount < n // 2:
a[pos] = 1
ocount += 1
notfulfilled += 1
elif notfulfilled > 1:
a[pos] = -1
ccount += 1
notfulfilled -= 1
elif a[pos] == -1:
if notfulfilled > 1:
notfulfilled -= 1
else:
print(":(")
quit()
if a[n - 1] == 1:
print(":(")
quit()
else:
a[n - 1] = -1
if sum(a) == 0:
for i in range(n):
if a[i] == 1:
print("(", end="")
elif a[i] == -1:
print(")", end="")
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR NUMBER VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER VAR NUMBER IF VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR STRING STRING IF VAR VAR NUMBER EXPR FUNC_CALL VAR STRING STRING EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input().strip())
ss = input().strip()
q = 0
for k in range(n):
if ss[k] == "(":
q += 1
ii = n // 2 - q
count = 0
s = ""
if n % 2 != 0:
print(":(")
elif ii < 0:
print(":(")
else:
for k in range(n):
if ss[k] == "?" and count < ii:
s = s + "("
count += 1
elif ss[k] == "?" and count >= ii:
s = s + ")"
elif ss[k] == "(":
s = s + "("
else:
s = s + ")"
c2 = 0
c3 = 0
for k in range(n - 1):
if s[k] == "(":
c2 += 1
else:
c3 += 1
if c3 >= c2:
print(":(")
break
else:
if s[n - 1] == "(":
c2 += 1
else:
c3 += 1
if c2 != c3:
print(":(")
else:
print(s)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR STRING IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR VAR ASSIGN VAR BIN_OP VAR STRING VAR NUMBER IF VAR VAR STRING VAR VAR ASSIGN VAR BIN_OP VAR STRING IF VAR VAR STRING ASSIGN VAR BIN_OP VAR STRING ASSIGN VAR BIN_OP VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING IF VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
l, r = 0, 0
for ch in s:
if ch == "(":
l += 1
elif ch == ")":
r += 1
mid = n // 2
if n % 2 == 1 or l > mid or r > mid:
print(":(")
exit(0)
res = ""
cnt = 0
for i in range(n):
ch = s[i]
if ch == "?":
if l < mid:
l += 1
ch = "("
else:
ch = ")"
res += ch
cnt += 1 if ch == "(" else -1
if cnt < 0 or cnt == 0 and i < n - 1:
print(":(")
exit(0)
print(res if cnt == 0 else ":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR STRING IF VAR VAR VAR NUMBER ASSIGN VAR STRING ASSIGN VAR STRING VAR VAR VAR VAR STRING NUMBER NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
if n == 0:
print("")
exit()
if s[0] == ")" or s[-1] == "(":
print(":(")
exit()
if n % 2 != 0:
print(":(")
exit()
q_marks_count = 0
l_brackets_count = 0
r_brackets_count = 0
for i in range(n):
if s[i] == "?":
q_marks_count += 1
elif s[i] == "(":
l_brackets_count += 1
else:
r_brackets_count += 1
need_l = max(r_brackets_count - l_brackets_count, 0)
need_r = max(l_brackets_count - r_brackets_count, 0)
q_marks_last = q_marks_count - max(need_l, need_r)
if q_marks_last < 0:
print(":(")
exit()
half = q_marks_last // 2
need_l += half
need_r += half
level = 0
for i in range(n):
if s[i] == "?":
if need_l > 0:
s[i] = "("
need_l -= 1
elif need_r > 0:
s[i] = ")"
need_r -= 1
if s[i] == "(":
level += 1
if s[i] == ")":
level -= 1
if level <= 0 and i != n - 1:
print(":(")
exit()
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
s = int(input())
inp = input()
open_count = inp.count("(")
close_count = inp.count(")")
open_new = s // 2 - open_count
close_new = s - open_count - close_count - open_new
bracket = list(inp)
for i in range(s):
if bracket[i] == "?":
if open_new > 0:
bracket[i] = "("
open_new -= 1
else:
bracket[i] = ")"
stck = [bracket[0]]
i = 1
while len(stck) > 0 and i < s:
if bracket[i] == "(":
stck.append(bracket[i])
else:
stck.pop()
i += 1
if len(stck) == 0 and i == s:
print("".join(bracket))
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR LIST VAR NUMBER ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
C0 = 0
C1 = 0
C2 = 0
L = []
flag = 0
cc0 = []
cc1 = []
for i in range(len(s) - 1, -1, -1):
L.append(s[i])
if s[i] == "(":
C0 += 1
elif s[i] == ")":
C1 += 1
else:
C2 += 1
cc0.append(C0)
cc1.append(C1)
cc0 = cc0[::-1]
cc1 = cc1[::-1]
L = L[::-1]
c0 = 0
c1 = 0
if L[0] == "?":
L[0] = "("
c0 += 1
elif L[0] == ")":
flag = 1
c1 += 1
else:
c0 += 1
if len(s) % 2 != 0:
flag = 1
for i in range(1, len(L) - 1):
if c0 <= c1:
flag = 1
break
if L[i] == ")":
c1 += 1
elif L[i] == "(":
c0 += 1
elif L[i] == "?":
if c0 > c1 + 1:
if c0 < len(s) // 2 and c0 + cc0[i] + 1 <= len(s) // 2:
L[i] = "("
c0 += 1
else:
L[i] = ")"
c1 += 1
else:
L[i] = "("
c0 += 1
if c0 <= c1:
flag = 1
if L[-1] == "(":
c0 += 1
flag = 1
elif L[-1] == ")":
c1 += 1
else:
L[-1] = ")"
c1 += 1
if c0 != c1:
flag = 1
if flag == 0:
L = "".join(L)
print(L)
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER STRING VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER STRING VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER STRING VAR NUMBER ASSIGN VAR NUMBER STRING VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
from sys import stdin, stdout
n = int(stdin.readline())
s = stdin.readline()
s = list(s)
B = True
if n & 1 or s[n - 1] == "(" or s[0] == ")" or s[n - 2] == "(":
print(":(")
else:
op = n // 2
cp = n // 2
aop = 0
acp = 0
cop = 0
if s[0] == "?":
s[0] = "("
if s[n - 1] == "?":
s[n - 1] = ")"
for i in s:
if i == "(":
aop += 1
elif i == ")":
acp += 1
rcp = cp - acp
rop = op - aop
if rcp < 0 or rop < 0:
print(":(")
B = False
else:
a = [(0) for i in range(n)]
for i in range(n):
if s[i] == "?":
if rop > 0:
a[i] = "("
rop -= 1
elif rcp > 0:
a[i] = ")"
rcp -= 1
else:
print(":(")
B = False
break
else:
a[i] = s[i]
if a[i] == "(":
cop += 1
else:
cop -= 1
if cop <= 0 and i != n - 1:
print(":(")
B = False
break
if B:
for i in a:
print(i, end="")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER STRING VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER STRING IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP VAR NUMBER STRING FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
import sys
sys_input = sys.stdin.readline
def si():
return sys_input().rstrip()
def ii():
return int(si())
def sti():
return si().split()
def iti():
return map(int, sti())
def sli():
return list(si())
def ili():
return list(iti())
n = ii()
s = sli()
def main():
if n == 2 and s[0] != ")" and s[1] != "(":
return "()"
if n % 2 == 1 or ")" in s[:2] or "(" in s[-2:]:
return ":("
s[0] = "("
s[1] = "("
s[-2] = ")"
s[-1] = ")"
p_num = n // 2
open_num = s.count("(")
close_num = s.count(")")
if open_num > p_num or close_num > p_num:
return ":("
result = s
cum_p_num = 0
q = n - open_num - close_num
for i, c in enumerate(s):
if c == "(":
cum_p_num += 1
open_num -= 1
if c == ")":
cum_p_num -= 1
close_num -= 1
if c == "?":
if (
cum_p_num + open_num + q > close_num
and cum_p_num + open_num == close_num + q
):
result[i] = ")"
cum_p_num -= 1
else:
result[i] = "("
cum_p_num += 1
q -= 1
if i < n - 1 and cum_p_num <= 0:
return ":("
return "".join(result)
print(main())
|
IMPORT ASSIGN VAR VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF VAR NUMBER VAR NUMBER STRING VAR NUMBER STRING RETURN STRING IF BIN_OP VAR NUMBER NUMBER STRING VAR NUMBER STRING VAR NUMBER RETURN STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING IF VAR VAR VAR VAR RETURN STRING ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR STRING IF BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER RETURN STRING RETURN FUNC_CALL STRING VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
res = []
x = 0
l = r = n // 2
l -= s.count("(")
r -= s.count(")")
y = r - l
f = s.count("?")
def check(d, b):
ans = x + d >= b
if d == 1:
ans &= l >= 1
else:
ans &= r >= 1
ans &= -f <= -(x + y + d) <= f
return ans
for i, c in enumerate(s):
if n & 1 or l < 0 or r < 0:
res = []
break
if c == "?":
if check(1, int(i != n - 1)):
l -= 1
res.append("(")
elif check(-1, int(i != n - 1)):
r -= 1
res.append(")")
else:
res = []
break
f -= 1
else:
res.append(c)
if res[-1] == "(":
x += 1
if c != "?":
y -= 1
else:
x -= 1
if c != "?":
y += 1
curr = 0
for i, c in enumerate(res):
if c == "(":
curr += 1
else:
curr -= 1
if curr < 0 or curr == 0 and i != n - 1:
res = []
break
if res:
print("".join(res))
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR STRING VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR STRING FUNC_DEF ASSIGN VAR BIN_OP VAR VAR VAR IF VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR RETURN VAR FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR LIST IF VAR STRING IF FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING IF FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR LIST VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER STRING VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR STRING VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR LIST IF VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = str(input())
s = list(s)
if n % 2 == 1:
print(":(")
quit()
a = b = 0
for i in s:
if i == "(":
a += 1
elif i == ")":
b += 1
a = n // 2 - a
b = n // 2 - b
for i in range(n):
if s[i] == "?":
if a == 0:
s[i] = ")"
b -= 1
else:
s[i] = "("
a -= 1
c = 0
for i in range(n):
if s[i] == "(":
c += 1
else:
c -= 1
if c < 0 or c == 0 and i != n - 1:
print(":(")
quit()
if c != 0:
print(":(")
quit()
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
if n % 2 == 1:
print(":(")
exit()
l, r = n // 2, n // 2
q = 0
for i in range(n):
if s[i] == "(":
l -= 1
elif s[i] == "?":
q += 1
else:
r += 1
if l < 0 or r < 0:
print(":(")
exit()
tmp = 0
for i in range(n):
if s[i] == "(":
tmp += 1
elif s[i] == "?":
if l > 0:
s[i] = "("
l -= 1
tmp += 1
else:
s[i] = ")"
tmp -= 1
r -= 1
else:
tmp -= 1
if i != n - 1 and tmp == 0 or tmp < 0 or l < 0 or r < 0:
print(":(")
exit()
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
a = []
for i in range(n):
if s[i] == "?":
a.append("(")
else:
a.append(s[i])
cnt = 0
ans = 0
for i in range(n):
if a[i] == "(":
cnt += 1
else:
cnt -= 1
ans = min(ans, cnt)
for i in range(n - 1, -1, -1):
if cnt > 0 and s[i] == "?":
cnt -= 2
a[i] = ")"
cnt = 0
b = []
for i in range(n):
if a[i] == "(":
cnt += 1
else:
cnt -= 1
b.append(cnt)
if len(b) == 1 or min(b[:-1]) <= 0 or b[-1] != 0:
ans = -1
if ans < 0:
print(":(")
else:
print("".join(a))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR NUMBER VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
if n % 2 != 0:
print(":(")
exit()
a = list(s)
xopen = a.count("(")
for i in range(len(a)):
if a[i] == "?":
if xopen < n // 2:
xopen += 1
a[i] = "("
else:
a[i] = ")"
diff = 0
for i in range(len(a)):
if a[i] == "(":
diff += 1
else:
diff -= 1
if diff <= 0 and i != len(a) - 1:
print(":(")
exit()
if diff != 0:
print(":(")
exit()
print("".join(a))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
seq = list(input())
ob = cb = res = 0
for i in range(0, n):
if seq[i] == "(":
ob = ob + 1
elif seq[i] == ")":
cb = cb + 1
for i in range(0, n):
if seq[i] == "?":
if ob < n / 2:
seq[i] = "("
ob = ob + 1
else:
seq[i] = ")"
cb = cb + 1
if seq[i] == "(":
res = res + 1
elif seq[i] == ")":
res = res - 1
if res < 1 and i != n - 1:
res = -1
break
if res == 0:
seq = "".join(seq)
print(seq)
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def solve():
if n % 2:
return ":("
c = 0
op = 0
for i in range(n):
if s[i] == "(":
op += 1
for i in range(n):
if op == n // 2:
break
if s[i] == "?":
s[i] = "("
op += 1
for i in range(n):
if i != 0 and c <= 0:
return ":("
if s[i] == "(":
c += 1
elif s[i] == ")":
c -= 1
else:
s[i] = ")"
c -= 1
if c == 0:
return "".join(s)
else:
return ":("
n = int(input())
s = list(input())
print(solve())
|
FUNC_DEF IF BIN_OP VAR NUMBER RETURN STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER RETURN STRING IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER RETURN FUNC_CALL STRING VAR RETURN STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
l = s.count("(")
cnt = 0
for i in range(n):
if s[i] == "?":
if l + l < n:
s[i] = "("
l += 1
else:
s[i] = ")"
for i in range(n):
if s[i] == "(":
cnt += 1
elif s[i] == ")":
cnt -= 1
if cnt < 0 or cnt != 0 and i == n - 1 or cnt == 0 and i != n - 1:
print(":(")
exit(0)
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF BIN_OP VAR VAR VAR ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def ex():
print(":(")
exit()
def check(q):
o, c = 0, 0
for i in q:
if i == "(":
o += 1
else:
if o == 0:
return 0
o -= 1
return 1
n = int(input())
s = list(input())
if n % 2:
ex()
o = s.count("(")
b = 0
for i in range(n):
if i == 0:
if s[i] == "?":
b = 1
s[i] = "("
o += 1
elif s[i] == "(":
b = 1
else:
b = -1
continue
if b == 0:
ex()
if s[i] == "?":
if o < n // 2:
s[i] = "("
b += 1
o += 1
else:
s[i] = ")"
b -= 1
elif s[i] == "(":
b += 1
else:
b -= 1
if b != 0 or not check(s):
ex()
print("".join(s))
|
FUNC_DEF EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR NUMBER RETURN NUMBER VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def is_true(s):
v = 0
for i in range(len(s)):
if s[i] == "(":
v = v + 1
else:
v = v - 1
if v < 0:
return False
if v == 0:
return True
return False
def solve(s):
n = len(s)
if n % 2 == 0:
if s[-1] != "(" and s[0] != ")":
s = list(s)
s[0] = "("
s[-1] = ")"
s = s[1:-1]
v = 0
nb = 0
for i in range(len(s)):
if s[i] == "?":
nb = nb + 1
if s[i] == "(":
v = v + 1
if s[i] == ")":
v = v - 1
if v > 0:
for i in range(len(s) - 1, -1, -1):
if s[i] == "?":
s[i] = ")"
nb = nb - 1
v = v - 1
if v == 0:
break
if v < 0:
for i in range(len(s)):
if s[i] == "?":
s[i] = "("
nb = nb - 1
v = v + 1
if v == 0:
break
if v == 0:
for i in range(len(s)):
if s[i] == "?":
if nb > 0:
s[i] = "("
nb = nb - 2
else:
s[i] = ")"
if is_true(s):
return "(" + "".join(s) + ")"
return ":("
n = int(input())
s = input()
print(solve(s))
|
FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER RETURN NUMBER IF VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR NUMBER STRING VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING IF FUNC_CALL VAR VAR RETURN BIN_OP BIN_OP STRING FUNC_CALL STRING VAR STRING RETURN STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
answer = []
left = s.count("(")
right = s.count(")")
l = r = 0
if n % 2 == 0:
for item in s:
if item == "(":
l += 1
answer.append("(")
elif item == ")":
r += 1
answer.append(")")
if r > l:
print(":(")
exit()
elif left < n // 2:
answer.append("(")
left += 1
l += 1
else:
answer.append(")")
right += 1
r += 1
l = r = 0
i = 0
for item in answer:
i += 1
if item == "(":
l += 1
else:
r += 1
if r > l or l == r and i != n and answer[i - 1] == ")" or l != r and i == n:
print(":(")
exit()
else:
print(*answer, sep="")
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER STRING VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
l = s.count("(")
r = s.count(")")
ans = "("
x = ":("
if n & 1 or l > n // 2 or r > n // 2:
print(x)
else:
l1 = n // 2 - l
r1 = n // 2 - r
l, r = 0, 0
ans = ""
i = 0
while i < len(s):
if s[i] == "?":
if l1 > 0:
ans += "("
l += 1
l1 -= 1
else:
ans += ")"
r += 1
r1 -= 1
elif s[i] == "(":
l += 1
ans += "("
else:
r += 1
ans += ")"
if r >= l and i != len(s) - 1:
ans = ":("
break
i += 1
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING IF BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER VAR STRING VAR NUMBER VAR NUMBER VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR STRING VAR NUMBER VAR STRING IF VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
l = 0
r = 0
for i in range(n):
if s[i] == "(":
r += 1
elif s[i] == ")":
l += 1
x = n - l - r
g = r - l
ee = (x - abs(g)) // 2
t = 0
for i in range(n):
if s[i] == "?":
if t < ee:
s[i] = "("
t += 1
elif t >= ee + abs(g):
s[i] = ")"
else:
if g > 0:
s[i] = ")"
else:
s[i] = "("
t += 1
ff = 0
for i in s[:-1]:
if i == "(":
ff += 1
else:
ff -= 1
if ff <= 0:
print(":(")
exit()
if s[-1] == ")":
ff -= 1
else:
ff += 1
if ff != 0:
print(":(")
exit()
print(*s, sep="")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR VAR ASSIGN VAR VAR STRING VAR NUMBER IF VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER STRING VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
from itertools import accumulate
from sys import exit
N = int(input())
S = input()
if S[0] == ")" or S[-1] == "(" or N % 2 == 1:
print(":(")
exit()
L = [0] * N
posi = 0
nega = 0
for i, s in enumerate(S):
if s == "(":
posi += 1
L[i] = 1
continue
elif s == ")":
nega += 1
L[i] = -1
continue
rem = N - posi - nega
sp = nega - posi
pi = (rem + sp) // 2
ni = (rem - sp) // 2
if pi < 0 or ni < 0:
print(":(")
exit()
LP = L.copy()
for i, a in enumerate(L):
if not a:
if pi > 0:
pi -= 1
LP[i] = 1
else:
LP[i] = -1
if all([(i > 0) for i in list(accumulate(LP))[:-1]]):
print("".join([("", "(", ")")[l] for l in LP]))
exit()
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR NUMBER STRING VAR NUMBER STRING BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR STRING VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR STRING VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING STRING STRING STRING VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
import sys
num = int(input())
raw = input()
inputs = [item for item in raw]
result = [item for item in raw]
if num % 2 == 0:
found = True
a = 0
b = 0
c = 0
for i in range(num):
if inputs[i] == "(":
a += 1
elif inputs[i] == ")":
b += 1
else:
c += 1
if (abs(a - b) + c) % 2 != 0 or a - b + c < 0 or b - a + c < 0:
print(":(")
else:
if a > b:
n = a - b
for i in range(num - 1, 0, -1):
if n == 0:
break
if inputs[i] == "?":
inputs[i] = ")"
n -= 1
elif a < b:
n = b - a
for i in range(0, num):
if n == 0:
break
if inputs[i] == "?":
inputs[i] = "("
n -= 1
n = (c - abs(a - b)) / 2
for i in range(0, num):
if n == 0:
break
if inputs[i] == "?":
inputs[i] = "("
n -= 1
n = (c - abs(a - b)) / 2
for i in range(num - 1, 0, -1):
if n == 0:
break
if inputs[i] == "?":
inputs[i] = ")"
n -= 1
sum = (inputs[0] == "(") - (inputs[0] == ")")
temp = 0
for i in range(1, num - 1):
sum = sum + (inputs[i] == "(") - (inputs[i] == ")")
if sum <= 0:
print(":(")
sys.exit(0)
print("".join(inputs))
else:
print(":(")
|
IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER STRING VAR NUMBER STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR STRING VAR VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
my_str = input()
strl = list(my_str)
num = [0] * n
if n % 2:
print(":(")
exit()
for i in range(n - 2, 0, -1):
num[i] = num[i + 1]
if strl[i] == "(":
num[i] = num[i] + 1
l = 0
s = 0
for i in range(n):
if strl[i] == "(":
l = l + 1
s = s + 1
elif strl[i] == ")":
if l > 0:
l = l - 1
if l == 0 and i != n - 1:
print(":(")
exit()
else:
print(":(")
exit()
else:
if s + num[i] < n // 2:
l = l + 1
s = s + 1
strl[i] = "("
else:
strl[i] = ")"
l = l - 1
if l <= 0 and i != n - 1:
print(":(")
exit()
if s != n // 2:
print(":(")
exit()
realstr = []
realstr = "".join(strl)
print(realstr)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def main():
n = int(input())
s = input()
mid = s[1 : len(s) - 1]
if n == 1:
print(":(")
return
correct = False
left = 0
right = 0
for i in mid:
if i == "(":
left += 1
elif i == ")":
right += 1
left = len(mid) // 2 - left
right = len(mid) // 2 - right
new = ""
for i in mid:
if i == "?":
if left != 0:
new += "("
left -= 1
elif right != 0:
new += ")"
right -= 1
else:
print(":(")
return
else:
new += i
left = 0
for i in new:
if i == "(":
left += 1
elif left > 0:
left -= 1
else:
print(":(")
return
if left > 0:
print(":(")
return
if (s[0] == "(" or s[0] == "?") and (s[-1] == ")" or s[-1] == "?"):
new = "(" + new
new += ")"
else:
print(":(")
return
print(new)
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR STRING FOR VAR VAR IF VAR STRING IF VAR NUMBER VAR STRING VAR NUMBER IF VAR NUMBER VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN IF VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN IF VAR NUMBER STRING VAR NUMBER STRING VAR NUMBER STRING VAR NUMBER STRING ASSIGN VAR BIN_OP STRING VAR VAR STRING EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
inp = input()
l, r = 0, 0
for i in range(n):
if inp[i] == "(":
l += 1
elif inp[i] == ")":
r += 1
lsaved = -1 * l + n // 2
rsaved = -1 * r + n // 2
res = []
for i in range(n):
if inp[i] == "?":
if lsaved:
res.append("(")
lsaved -= 1
elif rsaved:
res.append(")")
rsaved -= 1
else:
print(":(")
exit()
else:
res.append(inp[i])
ans = 0
rollingsum = 0
for i in range(n):
if i >= 1 and rollingsum == 0:
ans = 1
if res[i] == "(":
rollingsum += 1
else:
rollingsum -= 1
if rollingsum < 0:
ans = 1
if ans == 1 or rollingsum != 0:
print(":(")
exit()
print("".join(res))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
if n % 2 == 1:
print(":(")
else:
op = n // 2 - s.count("(")
for i in range(n):
if s[i] == "?":
if op > 0:
s[i] = "("
else:
s[i] = ")"
op -= 1
b = 0
for i in range(n):
if s[i] == ")":
b -= 1
else:
b += 1
if i != n - 1 and b <= 0:
print(":(")
break
else:
if b == 0:
print("".join(s))
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
open = 0
closed = 0
for i in range(n):
if s[i] == "(":
open += 1
elif s[i] == ")":
closed += 1
a = n / 2 - open
b = n / 2 - closed
ans = ""
openct = 0
closedct = 0
can = True
if n % 2 or open > n / 2:
can = False
can = False
if can:
for i in range(n):
if s[i] == "?":
if a:
openct += 1
ans += "("
a -= 1
else:
closedct += 1
ans += ")"
elif s[i] == "(":
ans += "("
openct += 1
elif s[i] == ")":
ans += ")"
closedct += 1
if i != n - 1:
if closedct >= openct:
can = False
break
if can:
print(ans)
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR VAR NUMBER VAR STRING VAR NUMBER VAR NUMBER VAR STRING IF VAR VAR STRING VAR STRING VAR NUMBER IF VAR VAR STRING VAR STRING VAR NUMBER IF VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
import sys
def rint():
return list(map(int, sys.stdin.readline().split()))
n = int(input())
s = input()
o = s.count("(")
c = s.count(")")
ans = []
cnt = 0
ok = 0
for i in range(n):
ans.append(s[i])
oc = o - c
if oc > 0:
for i in range(n - 1, -1, -1):
if oc == 0:
break
if ans[i] == "?":
ans[i] = ")"
oc -= 1
if oc != 0:
print(":(")
return
qc = n - o - c - (o - c)
if qc % 2:
print(":(")
return
q = 0
for i in range(n):
if q == qc:
break
if ans[i] == "?":
if q < qc // 2:
ans[i] = "("
else:
ans[i] = ")"
q += 1
cnt = 0
for i in range(n):
if ans[i] == "(":
cnt += 1
elif ans[i] == ")":
cnt -= 1
if cnt <= 0 and i != n - 1:
print(":(")
return
if cnt != 0:
print(":(")
return
print("".join(ans))
|
IMPORT FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN IF VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
if n % 2:
print(":(")
exit()
left = s.count("(")
right = s.count(")")
plus = 0
for i in range(n):
if s[i] == "?":
if left < n // 2:
s[i] = "("
left += 1
else:
s[i] = ")"
right += 1
if s[i] == "(":
plus += 1
else:
plus -= 1
if i != n - 1 and plus <= 0:
print(":(")
exit()
if left == n // 2 and right == n // 2:
print("".join(s))
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
if n % 2 == 1:
print(":(")
else:
opa = n // 2 - s.count("(")
k = 0
for i in range(n):
if s[i] == "(":
k += 1
elif s[i] == ")":
k -= 1
if i < n - 1 and k <= 0:
print(":(")
break
elif opa > 0:
opa -= 1
s[i] = "("
k += 1
else:
s[i] = ")"
k -= 1
if i < n - 1 and k <= 0:
print(":(")
break
else:
if k != 0:
print(":(")
else:
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
r = s[::-1]
cl, cr, tcr = 0, 0, 0
k = 0
p = ["("]
if n % 2 == 0:
if s[0] == ")" or s[n - 1] == "(":
print(":(")
else:
for i in range(1, n - 1):
if s[i] == ")":
cl += 1
if i - cl < cl:
k = 1
break
for i in range(1, n - 1):
if r[i] == "(":
cr += 1
if i - cr < cr:
k = 1
break
if k == 0:
if s[0] == "(":
c = s.count("(") - 1
else:
c = s.count("(")
i = 1
for i in range(1, n - 1):
if tcr + c >= n // 2 - 1:
break
if s[i] == "?":
p.append("(")
tcr += 1
else:
p.append(s[i])
for j in range(i, n):
if s[j] == "?":
p.append(")")
else:
p.append(s[j])
print("".join(p))
else:
print(":(")
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST STRING IF BIN_OP VAR NUMBER NUMBER IF VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR STRING VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR STRING VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR BIN_OP FUNC_CALL VAR STRING NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
l = list(s)
count, count1 = 0, 0
if n % 2 == 1:
print(":(")
exit()
for i in range(n):
if s[i] == "(":
count += 1
elif s[i] == ")":
count1 += 1
count = n // 2 - count
for j in range(n):
if s[j] == "?" and count > 0:
l[j] = "("
count -= 1
elif s[j] == "?":
l[j] = ")"
m = "".join(l)
count1 = 0
count = 0
if m[0] == ")" or m[n - 1] == "(":
print(":(")
exit()
for k in range(n):
if m[k] == "(":
count += 1
else:
count1 += 1
if count == count1 and count != n / 2:
print(":(")
exit()
if count == n / 2:
print(m)
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR STRING ASSIGN VAR VAR STRING ASSIGN VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def solve(s):
if len(s) == 1 or s[-1] == "(" or s[0] == ")" or len(s) % 2 == 1:
return ":("
ans = list(s)
ans[0] = "("
ans[-1] = ")"
left = []
star = []
for i in range(1, len(s) - 1):
if s[i] == "(":
left += [i]
elif s[i] == ")":
if left:
left.pop()
elif star:
p = star.pop()
ans[p] = "("
else:
return ":("
else:
star += [i]
while left and star and left[-1] < star[-1]:
left.pop()
p = star.pop()
ans[p] = ")"
if left:
return ":("
if len(star) % 2 == 1:
return ":("
for i in range(0, len(star) // 2):
p = star[i]
ans[p] = "("
for i in range(len(star) // 2, len(star)):
p = star[i]
ans[p] = ")"
return "".join(ans)
input()
s = input()
print(solve(s))
|
FUNC_DEF IF FUNC_CALL VAR VAR NUMBER VAR NUMBER STRING VAR NUMBER STRING BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER STRING ASSIGN VAR NUMBER STRING ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR STRING VAR LIST VAR IF VAR VAR STRING IF VAR EXPR FUNC_CALL VAR IF VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR STRING RETURN STRING VAR LIST VAR WHILE VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR STRING IF VAR RETURN STRING IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN STRING FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR STRING FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR STRING RETURN FUNC_CALL STRING VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
from sys import stdin, stdout
def get():
return stdin.readline().strip()
def getf():
return [int(i) for i in get().split()]
def put(a, end="\n"):
stdout.write(str(a) + end)
def putf(a, sep=" ", end="\n"):
stdout.write(sep.join([str(i) for i in a]) + end)
def bruh():
n = int(get())
s = [i for i in get()]
q = [0] * n
a = s.count("(")
b = s.count(")")
cnt = 0
for i in range(n):
if a * 2 < n:
if s[i] == "?":
s[i] = "("
a += 1
for i in range(n):
if b * 2 < n:
if s[i] == "?":
s[i] = ")"
b += 1
for i in range(n):
if s[i] == ")":
cnt -= 1
else:
cnt += 1
if cnt <= 0 and i != n - 1 or i == n - 1 and cnt != 0:
put(":(")
return 0
putf(s, "")
bruh()
|
FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_DEF STRING STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN NUMBER EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = input()
fail = ":("
if n % 2 != 0:
print(fail)
quit()
ocount = n // 2 - s.count("(")
ccount = n // 2 - s.count(")")
ans = []
balance = 0
for i, sym in enumerate(s):
if sym != "?":
ans.append(sym)
elif ocount > 0:
ocount -= 1
ans.append("(")
elif ccount > 0:
ccount -= 1
ans.append(")")
else:
raise Exception
x = ans[-1]
if x == "(":
balance += 1
else:
balance -= 1
if balance < 0:
print(fail)
quit()
elif i < n - 1 and balance == 0:
print(fail)
quit()
if balance != 0:
print(fail)
quit()
else:
print("".join(ans))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR STRING EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING VAR ASSIGN VAR VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
answer = [0] * n
st = 0
en = 0
for i in range(n):
if s[i] == "(":
st += 1
elif s[i] == ")":
en += 1
for i in range(n):
if s[i] == "?":
if st < n // 2:
s[i] = "("
st += 1
else:
s[i] = ")"
found = True
temp = [0] * n
if s[0] == "(":
temp[0] = 1
else:
found = False
i = 1
while i < n and found:
temp[i] = temp[i - 1]
if s[i] == "(":
temp[i] += 1
else:
temp[i] -= 1
if temp[i] <= 0 and i < n - 1:
found = False
break
i += 1
if found and temp[n - 1] == 0:
print("".join(s))
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR IF VAR NUMBER STRING ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR STRING VAR VAR NUMBER VAR VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
l = int(input())
s = input()
o = s.count("(")
c = s.count(")")
d = s.count("?")
numo, numc = 0, 0
if o < c:
numo += c - o
else:
numc += o - c
d -= numo
d -= numc
numo += d // 2
numc += d // 2
d -= 2 * d // 2
u = s.replace("?", "(", numo).replace("?", ")", numc)
p = 0
v = True
for c in u[: l - 1]:
if c == "(":
p += 1
else:
p -= 1
if p <= 0:
v = False
break
if u[l - 1] == "(":
p += 1
else:
p -= 1
if p != 0:
v = False
if v and d == 0:
print(u)
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING STRING VAR STRING STRING VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR BIN_OP VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
def main():
n = int(input())
arr = list(input())
if n & 1:
print(":(")
return 0
left = n // 2 - arr.count("(")
for i in range(n):
if arr[i] == "?":
if left:
arr[i] = "("
left -= 1
else:
arr[i] = ")"
if arr[0] == ")" or arr[-1] == "(":
print(":(")
return 0
left = 1
right = 0
for i in range(1, n):
if right == left:
print(":(")
return 0
if arr[i] == "(":
left += 1
else:
right += 1
if left != right:
print(":(")
return 0
print("".join(arr))
return 0
main()
|
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING RETURN NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR STRING RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR EXPR FUNC_CALL VAR STRING RETURN NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING RETURN NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR RETURN NUMBER EXPR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
if n % 2:
print(":(")
exit()
res = n // 2
for i in range(n):
if s[i] == "(":
res -= 1
for i in range(n):
if s[i] != "?":
continue
if res > 0:
s[i] = "("
res -= 1
else:
s[i] = ")"
c = 0
for i in range(n):
if s[i] == "(":
c += 1
else:
c -= 1
if c < 0 or i < n - 1 and c == 0:
print(":(")
exit()
if c:
print(":(")
else:
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
import sys
class Main:
def __init__(self):
self.buff = None
self.index = 0
def __next__(self):
if self.buff is None or self.index == len(self.buff):
self.buff = sys.stdin.readline().split()
self.index = 0
val = self.buff[self.index]
self.index += 1
return val
def next_int(self):
return int(next(self))
def cal(self, s):
if len(s) == 1:
return s[0]
if s[0] == 0:
return self.cal(s[1:])
v = 1
for c in s:
v *= c
return v
def solve(self):
n = self.next_int()
s = [x for x in next(self)]
w = s.count("?")
z = s.count("(")
y = n - w - z
ss = [x for x in s]
if ss[0] == "?":
ss[0] = "("
z += 1
if ss[-1] == "?":
ss[-1] = ")"
y += 1
flag = (
n % 2 == 0 and ss[0] != ")" and ss[-1] != "(" and z * 2 <= n and y * 2 <= n
)
c = 1
for i in range(1, n - 1):
if not flag:
break
if ss[i] == "(":
c += 1
elif ss[i] == ")":
c -= 1
elif z * 2 < n:
ss[i] = "("
c += 1
z += 1
else:
ss[i] = ")"
c -= 1
y -= 1
if c == 0:
flag = False
if not flag:
print(":(")
else:
print("".join(ss))
def __starting_point():
Main().solve()
__starting_point()
|
IMPORT CLASS_DEF FUNC_DEF ASSIGN VAR NONE ASSIGN VAR NUMBER FUNC_DEF IF VAR NONE VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER RETURN VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR NUMBER IF VAR NUMBER NUMBER RETURN FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR VAR IF VAR NUMBER STRING ASSIGN VAR NUMBER STRING VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER STRING VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER STRING VAR NUMBER STRING BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF BIN_OP VAR NUMBER VAR ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR FUNC_DEF EXPR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
N = int(input())
S = input()
l = 0
r = 0
for s in S:
if s == "(":
l += 1
elif s == ")":
r += 1
if N % 2 == 1:
ans = ":("
else:
lim = N // 2
ok = True
if l > lim or r > lim:
ok = False
ans = ""
for s in S:
if s == "(" or s == ")":
ans += s
elif l < lim:
ans += "("
l += 1
else:
ans += ")"
d = 0
for i in range(N - 1):
a = ans[i]
if a == "(":
d += 1
elif d > 1:
d -= 1
else:
ok = False
if not ok:
ans = ":("
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR VAR IF VAR STRING VAR STRING VAR VAR IF VAR VAR VAR STRING VAR NUMBER VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR IF VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
if n % 2 == 1 or s.count("(") > n // 2 or s.count(")") > n // 2:
print(":(")
exit()
ok = True
op, cl = 0, 0
for i in range(n):
if s[i] == ")":
cl += 1
else:
op += 1
if cl > op:
print(":(")
exit()
op, cl = 0, 0
for i in range(n - 1, -1, -1):
if s[i] == "(":
op += 1
else:
cl += 1
if cl < op:
print(":(")
exit()
al = s.count(")")
op, cl = 0, 0
for i in range(n - 1, 0, -1):
if s[i] == "(":
op += 1
elif s[i] == ")":
cl += 1
al -= 1
elif cl + al == n // 2:
s[i] = "("
op += 1
else:
s[i] = ")"
cl += 1
if op >= cl:
print(":(")
exit()
s[0] = "("
print("".join(s))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR STRING BIN_OP VAR NUMBER FUNC_CALL VAR STRING BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR ASSIGN VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
seq = list(input())
l = 0
r = 0
for c in seq:
if c == "(":
l += 1
elif c == ")":
r += 1
for i in range(len(seq) - 1, -1, -1):
if seq[i] == "?":
if r < n / 2:
seq[i] = ")"
r += 1
else:
seq[i] = "("
t = 0
broke = False
for i, c in enumerate(seq):
if c == "(":
t += 1
else:
t -= 1
if t <= 0:
if i != len(seq) - 1:
broke = True
break
if t == 0 and not broke:
print("".join(seq))
else:
print(":(")
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR STRING
|
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that $|s|$ as the length of string $s$. A strict prefix $s[1\dots l]$ $(1\leq l< |s|)$ of a string $s = s_1s_2\dots s_{|s|}$ is string $s_1s_2\dots s_l$. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string $s$ containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in $s$ independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
-----Input-----
The first line contains a single integer $|s|$ ($1\leq |s|\leq 3 \cdot 10^5$), the length of the string.
The second line contains a string $s$, containing only "(", ")" and "?".
-----Output-----
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
-----Examples-----
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
-----Note-----
It can be proved that there is no solution for the second sample, so print ":(".
|
n = int(input())
s = list(input())
def create_seq(n, s):
if n % 2 == 1:
return print(":(")
if s[0] == ")":
return print(":(")
open_limit = int(n / 2)
close_limit = int(n / 2)
for i in range(n):
if s[i] == "(":
open_limit -= 1
elif s[i] == ")":
close_limit -= 1
cnt_left = 0
cnt_right = 0
flg = 0
if (open_limit < 0) | (close_limit < 0):
flg = 1
for i in range(n):
if (cnt_left - cnt_right == 0) & (i > 0):
flg = 1
break
if (s[i] == "?") & (i == 0):
s[i] = "("
cnt_left += 1
open_limit -= 1
elif s[i] == "(":
cnt_left += 1
elif s[i] == ")":
cnt_right += 1
elif cnt_left - cnt_right >= 1:
if open_limit > 0:
s[i] = "("
cnt_left += 1
open_limit -= 1
elif close_limit > 0:
s[i] = ")"
cnt_right += 1
close_limit -= 1
elif cnt_left - cnt_right > 1:
if close_limit > 0:
s[i] = ")"
cnt_right += 1
close_limit -= 1
elif open_limit > 0:
s[i] = "("
cnt_left += 1
open_limit -= 1
if flg == 1:
return print(":(")
else:
return print("".join(s))
create_seq(n, s)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF BIN_OP VAR NUMBER NUMBER RETURN FUNC_CALL VAR STRING IF VAR NUMBER STRING RETURN FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR NUMBER IF BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER RETURN FUNC_CALL VAR STRING RETURN FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR VAR VAR
|
Elections in Berland are coming. There are only two candidates — Alice and Bob.
The main Berland TV channel plans to show political debates. There are $n$ people who want to take part in the debate as a spectator. Each person is described by their influence and political views. There are four kinds of political views: supporting none of candidates (this kind is denoted as "00"), supporting Alice but not Bob (this kind is denoted as "10"), supporting Bob but not Alice (this kind is denoted as "01"), supporting both candidates (this kind is denoted as "11").
The direction of the TV channel wants to invite some of these people to the debate. The set of invited spectators should satisfy three conditions: at least half of spectators support Alice (i.e. $2 \cdot a \ge m$, where $a$ is number of spectators supporting Alice and $m$ is the total number of spectators), at least half of spectators support Bob (i.e. $2 \cdot b \ge m$, where $b$ is number of spectators supporting Bob and $m$ is the total number of spectators), the total influence of spectators is maximal possible.
Help the TV channel direction to select such non-empty set of spectators, or tell that this is impossible.
-----Input-----
The first line contains integer $n$ ($1 \le n \le 4\cdot10^5$) — the number of people who want to take part in the debate as a spectator.
These people are described on the next $n$ lines. Each line describes a single person and contains the string $s_i$ and integer $a_i$ separated by space ($1 \le a_i \le 5000$), where $s_i$ denotes person's political views (possible values — "00", "10", "01", "11") and $a_i$ — the influence of the $i$-th person.
-----Output-----
Print a single integer — maximal possible total influence of a set of spectators so that at least half of them support Alice and at least half of them support Bob. If it is impossible print 0 instead.
-----Examples-----
Input
6
11 6
10 4
01 3
00 3
00 7
00 9
Output
22
Input
5
11 1
01 1
00 100
10 1
01 1
Output
103
Input
6
11 19
10 22
00 18
00 29
11 29
10 28
Output
105
Input
3
00 5000
00 5000
00 5000
Output
0
-----Note-----
In the first example $4$ spectators can be invited to maximize total influence: $1$, $2$, $3$ and $6$. Their political views are: "11", "10", "01" and "00". So in total $2$ out of $4$ spectators support Alice and $2$ out of $4$ spectators support Bob. The total influence is $6+4+3+9=22$.
In the second example the direction can select all the people except the $5$-th person.
In the third example the direction can select people with indices: $1$, $4$, $5$ and $6$.
In the fourth example it is impossible to select any non-empty set of spectators.
|
m = int(input())
dou = 0
fir = []
sec = []
non = []
d, f, s, n = 0, 0, 0, 0
for i in [0] * m:
a, b = list(map(int, input().split()))
if a == 11:
dou += b
d += 1
elif a == 1:
sec.append(b)
s += 1
elif a == 10:
fir.append(b)
f += 1
else:
non.append(b)
n += 1
fir.sort(reverse=True)
sec.sort(reverse=True)
if not d + f * s:
print(0)
quit()
else:
m = min(f, s)
ans = dou + sum(fir[:m]) + sum(sec[:m])
non = non + fir[m:] + sec[m:]
non.sort(reverse=True)
print(ans + sum(non[:d]))
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR VAR VAR NUMBER NUMBER NUMBER NUMBER FOR VAR BIN_OP LIST NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR
|
Elections in Berland are coming. There are only two candidates — Alice and Bob.
The main Berland TV channel plans to show political debates. There are $n$ people who want to take part in the debate as a spectator. Each person is described by their influence and political views. There are four kinds of political views: supporting none of candidates (this kind is denoted as "00"), supporting Alice but not Bob (this kind is denoted as "10"), supporting Bob but not Alice (this kind is denoted as "01"), supporting both candidates (this kind is denoted as "11").
The direction of the TV channel wants to invite some of these people to the debate. The set of invited spectators should satisfy three conditions: at least half of spectators support Alice (i.e. $2 \cdot a \ge m$, where $a$ is number of spectators supporting Alice and $m$ is the total number of spectators), at least half of spectators support Bob (i.e. $2 \cdot b \ge m$, where $b$ is number of spectators supporting Bob and $m$ is the total number of spectators), the total influence of spectators is maximal possible.
Help the TV channel direction to select such non-empty set of spectators, or tell that this is impossible.
-----Input-----
The first line contains integer $n$ ($1 \le n \le 4\cdot10^5$) — the number of people who want to take part in the debate as a spectator.
These people are described on the next $n$ lines. Each line describes a single person and contains the string $s_i$ and integer $a_i$ separated by space ($1 \le a_i \le 5000$), where $s_i$ denotes person's political views (possible values — "00", "10", "01", "11") and $a_i$ — the influence of the $i$-th person.
-----Output-----
Print a single integer — maximal possible total influence of a set of spectators so that at least half of them support Alice and at least half of them support Bob. If it is impossible print 0 instead.
-----Examples-----
Input
6
11 6
10 4
01 3
00 3
00 7
00 9
Output
22
Input
5
11 1
01 1
00 100
10 1
01 1
Output
103
Input
6
11 19
10 22
00 18
00 29
11 29
10 28
Output
105
Input
3
00 5000
00 5000
00 5000
Output
0
-----Note-----
In the first example $4$ spectators can be invited to maximize total influence: $1$, $2$, $3$ and $6$. Their political views are: "11", "10", "01" and "00". So in total $2$ out of $4$ spectators support Alice and $2$ out of $4$ spectators support Bob. The total influence is $6+4+3+9=22$.
In the second example the direction can select all the people except the $5$-th person.
In the third example the direction can select people with indices: $1$, $4$, $5$ and $6$.
In the fourth example it is impossible to select any non-empty set of spectators.
|
n = int(input())
a = []
b = []
c = []
d = []
for i in range(n):
e, f = list(map(int, input().split()))
if e == 11:
a.append(f)
elif e == 10:
b.append(f)
elif e == 1:
c.append(f)
elif e == 0:
d.append(f)
a.sort(reverse=True)
b.sort(reverse=True)
c.sort(reverse=True)
d.sort(reverse=True)
e = min(len(c), len(b))
f = a + b[:e] + c[:e]
g = d + b[e:] + c[e:]
g.sort(reverse=True)
g = g[: len(a)]
print(sum(f) + sum(g))
|
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 VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR
|
Elections in Berland are coming. There are only two candidates — Alice and Bob.
The main Berland TV channel plans to show political debates. There are $n$ people who want to take part in the debate as a spectator. Each person is described by their influence and political views. There are four kinds of political views: supporting none of candidates (this kind is denoted as "00"), supporting Alice but not Bob (this kind is denoted as "10"), supporting Bob but not Alice (this kind is denoted as "01"), supporting both candidates (this kind is denoted as "11").
The direction of the TV channel wants to invite some of these people to the debate. The set of invited spectators should satisfy three conditions: at least half of spectators support Alice (i.e. $2 \cdot a \ge m$, where $a$ is number of spectators supporting Alice and $m$ is the total number of spectators), at least half of spectators support Bob (i.e. $2 \cdot b \ge m$, where $b$ is number of spectators supporting Bob and $m$ is the total number of spectators), the total influence of spectators is maximal possible.
Help the TV channel direction to select such non-empty set of spectators, or tell that this is impossible.
-----Input-----
The first line contains integer $n$ ($1 \le n \le 4\cdot10^5$) — the number of people who want to take part in the debate as a spectator.
These people are described on the next $n$ lines. Each line describes a single person and contains the string $s_i$ and integer $a_i$ separated by space ($1 \le a_i \le 5000$), where $s_i$ denotes person's political views (possible values — "00", "10", "01", "11") and $a_i$ — the influence of the $i$-th person.
-----Output-----
Print a single integer — maximal possible total influence of a set of spectators so that at least half of them support Alice and at least half of them support Bob. If it is impossible print 0 instead.
-----Examples-----
Input
6
11 6
10 4
01 3
00 3
00 7
00 9
Output
22
Input
5
11 1
01 1
00 100
10 1
01 1
Output
103
Input
6
11 19
10 22
00 18
00 29
11 29
10 28
Output
105
Input
3
00 5000
00 5000
00 5000
Output
0
-----Note-----
In the first example $4$ spectators can be invited to maximize total influence: $1$, $2$, $3$ and $6$. Their political views are: "11", "10", "01" and "00". So in total $2$ out of $4$ spectators support Alice and $2$ out of $4$ spectators support Bob. The total influence is $6+4+3+9=22$.
In the second example the direction can select all the people except the $5$-th person.
In the third example the direction can select people with indices: $1$, $4$, $5$ and $6$.
In the fourth example it is impossible to select any non-empty set of spectators.
|
n = int(input())
tonumber = {"00": 0, "01": 1, "10": 2, "11": 3}
people = [[], [], [], []]
for i in range(n):
s, a = input().split()
s, a = tonumber[s], int(a)
people[s].append(a)
people[0].sort(reverse=True)
people[1].sort(reverse=True)
people[2].sort(reverse=True)
totalInfluence = sum(people[3])
totalPeople = len(people[3])
support = [len(people[3]), len(people[3])]
minLen = min(len(people[1]), len(people[2]))
for i in range(minLen):
totalInfluence += people[1][i] + people[2][i]
totalPeople += 2
support[0] += 1
support[1] += 1
longer = 0
longerIndex = 2
if minLen == len(people[1]):
longer = 0
longerIndex = 2
else:
longer = 1
longerIndex = 1
indices = [0, minLen]
allEmpty = [len(people[0]), len(people[longerIndex])]
while indices != allEmpty:
top = [-1, -1]
if indices[0] != allEmpty[0]:
top[0] = people[0][indices[0]]
if indices[1] != allEmpty[1]:
top[1] = people[longerIndex][indices[1]]
if top[0] > top[1]:
if (
support[longer] >= (totalPeople + 1) / 2
and support[not longer] >= (totalPeople + 1) / 2
):
totalInfluence += top[0]
totalPeople += 1
indices[0] += 1
else:
indices[0] += 1
elif support[not longer] >= (totalPeople + 1) / 2:
totalInfluence += top[1]
totalPeople += 1
support[longer] += 1
indices[1] += 1
else:
indices[1] += 1
if totalPeople != 0:
print(totalInfluence)
else:
print(0)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR DICT STRING STRING STRING STRING NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST LIST LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER VAR ASSIGN VAR LIST FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR VAR WHILE VAR VAR ASSIGN VAR LIST NUMBER NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER
|
Elections in Berland are coming. There are only two candidates — Alice and Bob.
The main Berland TV channel plans to show political debates. There are $n$ people who want to take part in the debate as a spectator. Each person is described by their influence and political views. There are four kinds of political views: supporting none of candidates (this kind is denoted as "00"), supporting Alice but not Bob (this kind is denoted as "10"), supporting Bob but not Alice (this kind is denoted as "01"), supporting both candidates (this kind is denoted as "11").
The direction of the TV channel wants to invite some of these people to the debate. The set of invited spectators should satisfy three conditions: at least half of spectators support Alice (i.e. $2 \cdot a \ge m$, where $a$ is number of spectators supporting Alice and $m$ is the total number of spectators), at least half of spectators support Bob (i.e. $2 \cdot b \ge m$, where $b$ is number of spectators supporting Bob and $m$ is the total number of spectators), the total influence of spectators is maximal possible.
Help the TV channel direction to select such non-empty set of spectators, or tell that this is impossible.
-----Input-----
The first line contains integer $n$ ($1 \le n \le 4\cdot10^5$) — the number of people who want to take part in the debate as a spectator.
These people are described on the next $n$ lines. Each line describes a single person and contains the string $s_i$ and integer $a_i$ separated by space ($1 \le a_i \le 5000$), where $s_i$ denotes person's political views (possible values — "00", "10", "01", "11") and $a_i$ — the influence of the $i$-th person.
-----Output-----
Print a single integer — maximal possible total influence of a set of spectators so that at least half of them support Alice and at least half of them support Bob. If it is impossible print 0 instead.
-----Examples-----
Input
6
11 6
10 4
01 3
00 3
00 7
00 9
Output
22
Input
5
11 1
01 1
00 100
10 1
01 1
Output
103
Input
6
11 19
10 22
00 18
00 29
11 29
10 28
Output
105
Input
3
00 5000
00 5000
00 5000
Output
0
-----Note-----
In the first example $4$ spectators can be invited to maximize total influence: $1$, $2$, $3$ and $6$. Their political views are: "11", "10", "01" and "00". So in total $2$ out of $4$ spectators support Alice and $2$ out of $4$ spectators support Bob. The total influence is $6+4+3+9=22$.
In the second example the direction can select all the people except the $5$-th person.
In the third example the direction can select people with indices: $1$, $4$, $5$ and $6$.
In the fourth example it is impossible to select any non-empty set of spectators.
|
def read():
return []
def main():
n = int(input())
d = [[], [], [], []]
for _ in range(n):
key, influence = input().split()
key = int(key, 2)
d[key].append(int(influence))
for key in range(4):
d[key].sort(reverse=True)
ans = sum(d[3])
additional = len(d[3])
container = []
if len(d[1]) < len(d[2]):
container = d[2]
elif len(d[1]) > len(d[2]):
container = d[1]
container_start = min(len(d[1]), len(d[2]))
zeros_start = 0
for a1, a2 in zip(d[1], d[2]):
ans += a1 + a2
while (container_start < len(container) or zeros_start < len(d[0])) and additional:
t1, t2 = 0, 0
if container_start < len(container):
t1 = container[container_start]
if zeros_start < len(d[0]):
t2 = d[0][zeros_start]
if t1 < t2:
ans += t2
zeros_start += 1
elif t1 >= t2 and t1 != 0:
ans += t1
container_start += 1
additional -= 1
print(ans)
def __starting_point():
main()
__starting_point()
|
FUNC_DEF RETURN LIST FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST LIST LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR NUMBER NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR IF VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_DEF EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR
|
Elections in Berland are coming. There are only two candidates — Alice and Bob.
The main Berland TV channel plans to show political debates. There are $n$ people who want to take part in the debate as a spectator. Each person is described by their influence and political views. There are four kinds of political views: supporting none of candidates (this kind is denoted as "00"), supporting Alice but not Bob (this kind is denoted as "10"), supporting Bob but not Alice (this kind is denoted as "01"), supporting both candidates (this kind is denoted as "11").
The direction of the TV channel wants to invite some of these people to the debate. The set of invited spectators should satisfy three conditions: at least half of spectators support Alice (i.e. $2 \cdot a \ge m$, where $a$ is number of spectators supporting Alice and $m$ is the total number of spectators), at least half of spectators support Bob (i.e. $2 \cdot b \ge m$, where $b$ is number of spectators supporting Bob and $m$ is the total number of spectators), the total influence of spectators is maximal possible.
Help the TV channel direction to select such non-empty set of spectators, or tell that this is impossible.
-----Input-----
The first line contains integer $n$ ($1 \le n \le 4\cdot10^5$) — the number of people who want to take part in the debate as a spectator.
These people are described on the next $n$ lines. Each line describes a single person and contains the string $s_i$ and integer $a_i$ separated by space ($1 \le a_i \le 5000$), where $s_i$ denotes person's political views (possible values — "00", "10", "01", "11") and $a_i$ — the influence of the $i$-th person.
-----Output-----
Print a single integer — maximal possible total influence of a set of spectators so that at least half of them support Alice and at least half of them support Bob. If it is impossible print 0 instead.
-----Examples-----
Input
6
11 6
10 4
01 3
00 3
00 7
00 9
Output
22
Input
5
11 1
01 1
00 100
10 1
01 1
Output
103
Input
6
11 19
10 22
00 18
00 29
11 29
10 28
Output
105
Input
3
00 5000
00 5000
00 5000
Output
0
-----Note-----
In the first example $4$ spectators can be invited to maximize total influence: $1$, $2$, $3$ and $6$. Their political views are: "11", "10", "01" and "00". So in total $2$ out of $4$ spectators support Alice and $2$ out of $4$ spectators support Bob. The total influence is $6+4+3+9=22$.
In the second example the direction can select all the people except the $5$-th person.
In the third example the direction can select people with indices: $1$, $4$, $5$ and $6$.
In the fourth example it is impossible to select any non-empty set of spectators.
|
n = int(input())
t_11 = []
t_10 = []
t_01 = []
t_00 = []
for x in range(n):
tmp = list(input().split())
supp = str(tmp[0])
m = int(tmp[1])
t_tmp = [supp, m]
if supp == "11":
t_11.append(t_tmp)
elif supp == "01":
t_01.append(t_tmp)
elif supp == "10":
t_10.append(t_tmp)
else:
t_00.append(t_tmp)
t_00 = sorted(t_00, key=lambda x: x[1])
t_10 = sorted(t_10, key=lambda x: x[1])
t_01 = sorted(t_01, key=lambda x: x[1])
t_11 = sorted(t_11, key=lambda x: x[1])
pop_a = 0
pop_b = 0
tot_inf = 0
pop_tot = 0
l_t_11 = len(t_11) - 1
for x in range(l_t_11, -1, -1):
pop_a += 1
pop_b += 1
pop_tot += 1
tot_inf += t_11[-1][1]
del t_11[-1]
for x in range(min(len(t_10), len(t_01)) - 1, -1, -1):
pop_b += 1
pop_a += 1
pop_tot += 2
tot_inf += t_01[-1][1] + t_10[-1][1]
del t_01[-1]
del t_10[-1]
tmp_inf_1 = 0
tmp_inf_2 = 0
tmp_inf_3 = 0
tmp_t1 = []
tmp_t2 = []
if len(t_10) != 0:
for x in t_10:
tmp_t1.append(x)
for x in t_00:
tmp_t1.append(x)
tmp_t1 = sorted(tmp_t1, key=lambda x: x[1])
while True:
if (
len(tmp_t1) == 0
or pop_a * 2 == pop_tot
or pop_b * 2 == pop_a
or pop_b * 2 == pop_tot
):
break
tot_inf += tmp_t1[-1][1]
if tmp_t1[-1][0] == "10":
pop_a += 1
pop_tot += 1
else:
pop_tot += 1
del tmp_t1[-1]
print(tot_inf)
elif len(t_01) != 0:
for x in t_01:
tmp_t2.append(x)
for x in t_00:
tmp_t2.append(x)
tmp_t2 = sorted(tmp_t2, key=lambda x: x[1])
while True:
if (
len(tmp_t2) == 0
or pop_b * 2 == pop_tot
or pop_a * 2 == pop_b
or pop_a * 2 == pop_tot
):
break
tot_inf += tmp_t2[-1][1]
if tmp_t2[-1][0] == "01":
pop_b += 1
pop_tot += 1
else:
pop_tot += 1
del tmp_t2[-1]
print(tot_inf)
elif len(t_00) == 0:
print(tot_inf)
return
else:
while True:
if len(t_00) == 0 or pop_a * 2 == pop_tot or pop_b * 2 == pop_tot:
break
tot_inf += t_00[-1][1]
pop_tot += 1
del t_00[-1]
print(tot_inf)
|
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 FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST VAR VAR IF VAR STRING EXPR FUNC_CALL VAR VAR IF VAR STRING EXPR FUNC_CALL VAR VAR IF VAR STRING EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST IF FUNC_CALL VAR VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER WHILE NUMBER IF FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER NUMBER IF VAR NUMBER NUMBER STRING VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER WHILE NUMBER IF FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER NUMBER IF VAR NUMBER NUMBER STRING VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN WHILE NUMBER IF FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
|
Elections in Berland are coming. There are only two candidates — Alice and Bob.
The main Berland TV channel plans to show political debates. There are $n$ people who want to take part in the debate as a spectator. Each person is described by their influence and political views. There are four kinds of political views: supporting none of candidates (this kind is denoted as "00"), supporting Alice but not Bob (this kind is denoted as "10"), supporting Bob but not Alice (this kind is denoted as "01"), supporting both candidates (this kind is denoted as "11").
The direction of the TV channel wants to invite some of these people to the debate. The set of invited spectators should satisfy three conditions: at least half of spectators support Alice (i.e. $2 \cdot a \ge m$, where $a$ is number of spectators supporting Alice and $m$ is the total number of spectators), at least half of spectators support Bob (i.e. $2 \cdot b \ge m$, where $b$ is number of spectators supporting Bob and $m$ is the total number of spectators), the total influence of spectators is maximal possible.
Help the TV channel direction to select such non-empty set of spectators, or tell that this is impossible.
-----Input-----
The first line contains integer $n$ ($1 \le n \le 4\cdot10^5$) — the number of people who want to take part in the debate as a spectator.
These people are described on the next $n$ lines. Each line describes a single person and contains the string $s_i$ and integer $a_i$ separated by space ($1 \le a_i \le 5000$), where $s_i$ denotes person's political views (possible values — "00", "10", "01", "11") and $a_i$ — the influence of the $i$-th person.
-----Output-----
Print a single integer — maximal possible total influence of a set of spectators so that at least half of them support Alice and at least half of them support Bob. If it is impossible print 0 instead.
-----Examples-----
Input
6
11 6
10 4
01 3
00 3
00 7
00 9
Output
22
Input
5
11 1
01 1
00 100
10 1
01 1
Output
103
Input
6
11 19
10 22
00 18
00 29
11 29
10 28
Output
105
Input
3
00 5000
00 5000
00 5000
Output
0
-----Note-----
In the first example $4$ spectators can be invited to maximize total influence: $1$, $2$, $3$ and $6$. Their political views are: "11", "10", "01" and "00". So in total $2$ out of $4$ spectators support Alice and $2$ out of $4$ spectators support Bob. The total influence is $6+4+3+9=22$.
In the second example the direction can select all the people except the $5$-th person.
In the third example the direction can select people with indices: $1$, $4$, $5$ and $6$.
In the fourth example it is impossible to select any non-empty set of spectators.
|
d = {"00": [], "01": [], "10": [], "11": []}
n = int(input())
for i in range(n):
x, p = input().split()
d[x].append(int(p))
for key in d.keys():
d[key].sort(reverse=True)
t = "01" if len(d["01"]) <= len(d["10"]) else "10"
t1 = "01" if t == "10" else "10"
ans = 0
ans += sum(d["11"]) + sum(d[t]) + sum(d[t1][: len(d[t])])
e = len(d["11"])
l1 = sorted(d[t1][len(d[t]) :] + d["00"], reverse=True)
ans += sum(l1[:e])
print(ans)
|
ASSIGN VAR DICT STRING STRING STRING STRING LIST LIST LIST LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR STRING FUNC_CALL VAR VAR STRING STRING STRING ASSIGN VAR VAR STRING STRING STRING ASSIGN VAR NUMBER VAR BIN_OP BIN_OP FUNC_CALL VAR VAR STRING FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR STRING ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR VAR STRING NUMBER VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR
|
Elections in Berland are coming. There are only two candidates — Alice and Bob.
The main Berland TV channel plans to show political debates. There are $n$ people who want to take part in the debate as a spectator. Each person is described by their influence and political views. There are four kinds of political views: supporting none of candidates (this kind is denoted as "00"), supporting Alice but not Bob (this kind is denoted as "10"), supporting Bob but not Alice (this kind is denoted as "01"), supporting both candidates (this kind is denoted as "11").
The direction of the TV channel wants to invite some of these people to the debate. The set of invited spectators should satisfy three conditions: at least half of spectators support Alice (i.e. $2 \cdot a \ge m$, where $a$ is number of spectators supporting Alice and $m$ is the total number of spectators), at least half of spectators support Bob (i.e. $2 \cdot b \ge m$, where $b$ is number of spectators supporting Bob and $m$ is the total number of spectators), the total influence of spectators is maximal possible.
Help the TV channel direction to select such non-empty set of spectators, or tell that this is impossible.
-----Input-----
The first line contains integer $n$ ($1 \le n \le 4\cdot10^5$) — the number of people who want to take part in the debate as a spectator.
These people are described on the next $n$ lines. Each line describes a single person and contains the string $s_i$ and integer $a_i$ separated by space ($1 \le a_i \le 5000$), where $s_i$ denotes person's political views (possible values — "00", "10", "01", "11") and $a_i$ — the influence of the $i$-th person.
-----Output-----
Print a single integer — maximal possible total influence of a set of spectators so that at least half of them support Alice and at least half of them support Bob. If it is impossible print 0 instead.
-----Examples-----
Input
6
11 6
10 4
01 3
00 3
00 7
00 9
Output
22
Input
5
11 1
01 1
00 100
10 1
01 1
Output
103
Input
6
11 19
10 22
00 18
00 29
11 29
10 28
Output
105
Input
3
00 5000
00 5000
00 5000
Output
0
-----Note-----
In the first example $4$ spectators can be invited to maximize total influence: $1$, $2$, $3$ and $6$. Their political views are: "11", "10", "01" and "00". So in total $2$ out of $4$ spectators support Alice and $2$ out of $4$ spectators support Bob. The total influence is $6+4+3+9=22$.
In the second example the direction can select all the people except the $5$-th person.
In the third example the direction can select people with indices: $1$, $4$, $5$ and $6$.
In the fourth example it is impossible to select any non-empty set of spectators.
|
n = int(input())
su0 = []
su1 = []
su00 = []
su11 = []
ans = 0
for i in range(n):
sup, val = map(int, input().split())
if sup == 0:
su00.append(val)
elif sup == 11:
su11.append(val)
elif sup == 1:
su0.append(val)
else:
su1.append(val)
su0.sort()
su1.sort()
su11.sort()
su00.sort()
while len(su1) != 0 and len(su0) != 0:
ans += su1[len(su1) - 1] + su0[len(su0) - 1]
su1.pop()
su0.pop()
while len(su11) != 0 and len(su00) != 0:
if len(su1) != 0:
k = 1
elif len(su0) != 0:
k = 0
else:
k = 2
if k == 0:
if su00[len(su00) - 1] > su0[len(su0) - 1]:
ans += su11[len(su11) - 1] + su00[len(su00) - 1]
su11.pop()
su00.pop()
else:
ans += su11[len(su11) - 1] + su0[len(su0) - 1]
su11.pop()
su0.pop()
elif k == 1:
if su00[len(su00) - 1] > su1[len(su1) - 1]:
ans += su11[len(su11) - 1] + su00[len(su00) - 1]
su11.pop()
su00.pop()
else:
ans += su11[len(su11) - 1] + su1[len(su1) - 1]
su11.pop()
su1.pop()
else:
ans += su11[len(su11) - 1] + su00[len(su00) - 1]
su11.pop()
su00.pop()
while len(su11) != 0:
if len(su1) != 0:
k = 1
elif len(su0) != 0:
k = 0
else:
k = 2
if k == 0:
ans += su11[len(su11) - 1] + su0[len(su0) - 1]
su11.pop()
su0.pop()
elif k == 1:
ans += su11[len(su11) - 1] + su1[len(su1) - 1]
su11.pop()
su1.pop()
else:
ans += su11[len(su11) - 1]
su11.pop()
print(ans)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR WHILE FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR WHILE FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR WHILE FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
|
Elections in Berland are coming. There are only two candidates — Alice and Bob.
The main Berland TV channel plans to show political debates. There are $n$ people who want to take part in the debate as a spectator. Each person is described by their influence and political views. There are four kinds of political views: supporting none of candidates (this kind is denoted as "00"), supporting Alice but not Bob (this kind is denoted as "10"), supporting Bob but not Alice (this kind is denoted as "01"), supporting both candidates (this kind is denoted as "11").
The direction of the TV channel wants to invite some of these people to the debate. The set of invited spectators should satisfy three conditions: at least half of spectators support Alice (i.e. $2 \cdot a \ge m$, where $a$ is number of spectators supporting Alice and $m$ is the total number of spectators), at least half of spectators support Bob (i.e. $2 \cdot b \ge m$, where $b$ is number of spectators supporting Bob and $m$ is the total number of spectators), the total influence of spectators is maximal possible.
Help the TV channel direction to select such non-empty set of spectators, or tell that this is impossible.
-----Input-----
The first line contains integer $n$ ($1 \le n \le 4\cdot10^5$) — the number of people who want to take part in the debate as a spectator.
These people are described on the next $n$ lines. Each line describes a single person and contains the string $s_i$ and integer $a_i$ separated by space ($1 \le a_i \le 5000$), where $s_i$ denotes person's political views (possible values — "00", "10", "01", "11") and $a_i$ — the influence of the $i$-th person.
-----Output-----
Print a single integer — maximal possible total influence of a set of spectators so that at least half of them support Alice and at least half of them support Bob. If it is impossible print 0 instead.
-----Examples-----
Input
6
11 6
10 4
01 3
00 3
00 7
00 9
Output
22
Input
5
11 1
01 1
00 100
10 1
01 1
Output
103
Input
6
11 19
10 22
00 18
00 29
11 29
10 28
Output
105
Input
3
00 5000
00 5000
00 5000
Output
0
-----Note-----
In the first example $4$ spectators can be invited to maximize total influence: $1$, $2$, $3$ and $6$. Their political views are: "11", "10", "01" and "00". So in total $2$ out of $4$ spectators support Alice and $2$ out of $4$ spectators support Bob. The total influence is $6+4+3+9=22$.
In the second example the direction can select all the people except the $5$-th person.
In the third example the direction can select people with indices: $1$, $4$, $5$ and $6$.
In the fourth example it is impossible to select any non-empty set of spectators.
|
n = int(input())
influences = [[], [], [], []]
for i in range(0, n):
[support, influence] = [x for x in input().split()]
support = int(support, 2)
influences[support].append(int(influence))
for i in range(0, 4):
influences[i] = sorted(influences[i])
influences[i].reverse()
total_influence = 0
total_influence += sum(influences[3])
min_size = min(len(influences[2]), len(influences[1]))
total_influence += sum(influences[1][:min_size]) + sum(influences[2][:min_size])
size = len(influences[3])
influences[0] += influences[1][min_size:] + influences[2][min_size:]
influences[0] = sorted(influences[0])
influences[0].reverse()
total_influence += sum(influences[0][:size])
print(total_influence)
|
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST LIST LIST LIST LIST FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN LIST VAR VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR
|
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