优化 Python 代码以在超过时间限制时读取各种输入

Question

下面是我的 python 代码，我希望能够针对不同的输入优化我的代码，并通过我当前代码不断收到的时间限制错误。一些示例输入针对 3 个案例 n 每个案例都会有一行 space 如此分隔。

我已经尝试创建不同的方式来获取输入，即使对于大值的输入也是如此，但我无法通过不同测试用例的时间限制错误。如果有人可以告诉我我的代码在哪里，它会减慢这个过程，这将非常有帮助

 Sample tests:
 3
 3 2 1 1
 2 1 1 1
 5 2 1 1

n = int(input())
if 1<= n <= 100000:
counter = 0
while counter < n:
    i = 1
    a = list(map(int, input().split(' ')))
    D = a[0]
    d = a[1]
    P = a[2]
    Q = a[3]
    if (1 <= d <= D <= 1000000) and d <= D <= 100:
        if 1 <= P and Q <= 1000000:
            days = 1
            while i <= D:
                if i % d == 0:
                    if days == 1:
                        Production = P + Q
                        Total_money = d*P
                        days+=1
                    elif days > 1:
                        Total_money+= Production*d
                        days+= 1
                elif i%d == 1 and i == D:
                    if days <= 2:
                        Total_money+= Production
                    else:
                        Total_money+= Production + Q
                i+= 1
            counter+= 1
            print(Total_money)

Answer 1

以上代码可以改进如下。注意：条件不是必需的，因为它们通常用于提醒您必须应对的值范围。

def calc_total_money(D, d, P, Q):
    '''

     Calculates the total money 

     The intervals are made up of:

    Days
    [   d days   |   d days   |  d days |  .... | d days | r days]

    Amounts per Day
    [P           |  P+Q       | P + 2*Q | P + 3*Q ...             ]

    '''
    intervals = D // d                    # Number of whole intervals  of d days
    r = D - d*intervals                   # Number of days in last interval (i.e. partial interval)

    if intervals == 0:
        # No whole intervals
        return P*r
    else:
        # intervals = number of intervals of d days (i.e. whole intervals)
        # Amount in whole intervals are:
        # P, P+Q, P + 2*Q, P + 3*Q, ... P + (intervals-1)*Q
        # This equals: P*intervals + Q*intervals*(intervals - 1)//2
        # Since each interval is d days we have amount from whole interval of:
        # amount_whole_intervals = (P*intervals + Q*intervals*(intervals - 1)//2)*d
        #
        # Amount per day in last partial interval:
        #  P + intervals*Q
        # There are r days in last partial interval, so amount in last partial interval is:
        # last_partial_amount = (P + intervals*Q)*r
        #
        # Total = amount_whole_intervals + last_partial_amount
        return  (P*intervals + Q*intervals*(intervals - 1)//2)*d + (P + intervals*Q)*r

    
for _ in range(int(input())):
    D, d, P, Q = map(int, input().split(' '))
    print(calc_total_money(D, d, P, Q))

测试

输入

输出（与OP代码相同）

4
3
9

优化 Python 代码以在超过时间限制时读取各种输入

Optimize Python Code to read various inputs as time limit exceeded

python

if-statement

time-complexity

while-loop