思路:模拟题。直接用一个cnt变量,表示模拟栈括号匹配时的深度,然后每次成功匹配一个')'时,更新答案,最终输出结果即可
代码:
import sys
input = lambda: sys.stdin.readline().strip()
import math
inf = 10 ** 18
def I():
return input()
def II():
return int(input())
def MII():
return map(int, input().split())
def GMI():
return map(lambda x: int(x) - 1, input().split())
def LI():
return input().split()
def LII():
return list(map(int, input().split()))
def LFI():
return list(map(float, input().split()))
fmax = lambda x, y: x if x > y else y
fmin = lambda x, y: x if x < y else y
isqrt = lambda x: int(math.sqrt(x))
'''
'''
def solve():
s = I()
ans = cnt = 0
for c in s:
if c == '(':
cnt += 1
else:
ans = fmax(ans, cnt)
cnt -= 1
print(ans)
t = 1
# t = II()
for _ in range(t):
solve()

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