设计函数求一元多项式的导数。（注：x_n（n为整数）的一阶导数为nxⁿ⁻¹。）

输入格式:

以指数递降方式输入多项式非零项系数和指数（绝对值均为不超过 1000 的整数）。数字间以空格分隔。

输出格式:

以与输入相同的格式输出导数多项式非零项的系数和指数。数字间以空格分隔，但结尾不能有多余空格。注意“零多项式”的指数和系数都是 0，但是表示为 0 0。

输入样例:

3 4 -5 2 6 1 -2 0

输出样例:

12 3 -10 1 6 0

编程语言：Python

解题思想：

1.因为题目的输入是递减的多项式，所以一旦发现某个多项式指数小于等于0就要结束求导，只输出之前的结果

2.创建两个列表存放系数、指数，再创建一个列表存放求导后的多项式结果

3.对具体的情况进行判断

时间复杂度：O(N)

代码如下：

Number = list(map(int, input().split()))
 
Coefficient, Index = list(), list()  # 存放多项式的系数、指数
Derivative = list()
 
for i in range(len(Number)):
    if i % 2 == 0:
        Coefficient.append(Number[i])  # 存放系数
    else:
        Index.append(Number[i])  # 存放指数
 
for i in range(len(Coefficient)):
    if Coefficient[i] != 0 and Index[i] != 0:
        Coefficient[i] *= Index[i]
        Index[i] -= 1
        Derivative.append([Coefficient[i], Index[i]])  # 存放求导后的多项式
    elif Coefficient[i] == 0:
        Coefficient[i], Index[i] = 0, 0
        Derivative.append([Coefficient[i], Index[i]])  # 存放求导后的多项式
 
if len(Derivative) == 0:  # 首项的指数就小于等于0，导致后续多项式不再求导，这种情况需要输出0 0
    print(0, 0)
else:
    for i in range(len(Derivative)):
        if i != len(Derivative) - 1:
            print(Derivative[i][0], Derivative[i][1], end=' ')
        else:
            print(Derivative[i][0], Derivative[i][1])

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PAT乙级:1010 一元多项式求导