题解 | #四则运算#

import sys


def cal(inp: str):
    # 先算括号，把所有括号都清除，从大括号到小括号
    for brace in ("{", "[", "("):
        if brace in inp:
            # 找到当前的第一对括号
            start, end = find_brace(inp, brace)
			# 递归计算括号内的结果，然后重新组装入参
            inp = inp[:start] + str(cal(inp[start + 1 : end])) + inp[end + 1 :]
			# 递归地得到结果
            return cal(inp)

    # 把去掉括号的字符串拆解成符号和数字，以列表形式存储，方便处理
    stack = devide_inp(inp)
	# 先计算乘除法，因为优先级更高
    stack = cal_multi(stack)
	# 最后计算加减法
    res = cal_plus(stack)
    return int(res)


def find_brace(inp, brace):
    """找到字符串inp的brace括号的第一个完整的括号序号"""
    start, end = None, None
	# 因为可能有重叠的括号，需要找到完整括号内的内容
    brace_stack = []
    brace_map = {"{": "}", "[": "]", "(": ")"}
    for i, each in enumerate(inp):
        if each == brace:
            brace_stack.append(brace)
            if start is None:
                start = i
        elif each == brace_map[brace]:
            brace_stack.pop()
            if not brace_stack:
                end = i
                break

    return start, end


def devide_inp(inp: str):
    """把输入的字符串拆解成符号与数字"""
    start, end = 0, 0
    stack = []
    for i, each in enumerate(inp):
        if each not in ("+", "-", "*", "/"):
            continue
        end = i
        # 拆分的时候注意负数的计算，如果两种符号连着的，就把负号也带上
        if each == "-" and end - start == 0:
            continue

        stack.append(float(inp[start:end]))
        stack.append(each)
        start = i + 1

    stack.append(float(inp[start:]))
    return stack


def cal_multi(stack: list):
    # 处理乘除的计算，直到再也找不到
	# 从后往前处理，可以避免stack处理后，其后的序号出现变化的问题
    for i in range(len(stack)-2, 0, -2):
        if stack[i] not in ("*", "/"):
            continue

        if stack[i] == "*":
            stack[i-1] *= stack[i+1]
            stack.pop(i)
            stack.pop(i)
        else:
            stack[i-1] *= stack[i+1]
            stack.pop(i)
            stack.pop(i)

    return stack


def cal_plus(stack: list):
    # 处理加减
    while len(stack) > 1:
        if stack[1] == "+":
            stack[0] += stack[2]
            stack.pop(1)
            stack.pop(1)
        elif stack[1] == "-":
            stack[0] -= stack[2]
            stack.pop(1)
            stack.pop(1)
    return stack[0]




for line in sys.stdin:
    print(cal(line))
	
	# 以下用Python的eval即可解决
    # line = line.replace("[", "(")
    # line = line.replace("]", ")")
    # line = line.replace("{", "(")
    # line = line.replace("}", ")")
    # print(int(eval(line)))