This time, you are supposed to find $A\times B$ where $A$ and $B$ are two polynomials.
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
$K$ $N_1 a_{N_1} N_2 a_{N_2} \cdots N_K a_{N_K}$
where $K$ is the number of nonzero terms in the polynomial, $N_i$ and $a_{N_i} (i=1,2,\cdots,K)$ are the exponents and coefficients, respectively. It is given that $1\leqslant K\leqslant 10$，$0\leqslant N_K < \cdots < N_2 < N_1 \leqslant 1000$.
For each test case you should output the product of $A$ and $B$ in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.
2 1 2.4 0 3.2
2 2 1.5 1 0.5
3 3 3.6 2 6.0 1 1.6
本题 25 分，是比较简单的模拟题。在输入时，Python可以不用考虑给定的$K$，直接用数组切片隔位取指数列表、系数列表，再用zip()构造字典。
重点需要考虑的是如何将多项式的指数和对应系数映射。C/C++可以用数组下标作为指数，数组浮点数值为对应系数。这里Python使用字典比较方便，而且考虑到两个多项式的指数项不一定相同，在求和时要指数项系数对应相加，Python可以使用集合的求并集|运算符实现。
注意，测试点6是非零系数项个数，也就是$K$为0的情况，这时候只输出0。如果有多余空格会出现“格式错误”。
from itertools import product
ins1 = input()
ins2 = input()
a = zip( map(int, ins1.split()[1::2]), map(float, ins1.split()[2::2]) )
b = zip( map(int, ins2.split()[1::2]), map(float, ins2.split()[2::2]) )
poly = {}
prod = map(lambda x: (x[0][0]+x[1][0], x[0][1]*x[1][1]), product(a, b))
for exp, coef in prod:
poly[exp] = poly.get(exp, 0.0) + coef
poly = list( filter(lambda x:x[1]!=0.0, sorted(poly.items(), reverse=True)) )
line = ' '.join(map(lambda x: '{:d} {:.1f}'.format(*x), poly))
k = len(poly)
if k:
print(k, line)
else:
print(0)
A1002多项式加法的解析：https://column.mowchan.me/post/2021/pat-a1002.html