PAT (Advanced Level) Practice 1009 Product of Polynomials（Python）

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题目

This time, you are supposed to find A×B where A and B are two polynomials.

Input Specification:
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
$K\;N_1\;a_{N1}\;N_2\;a_{N2}\dots\;N_k\;a_{Nk}$
where K is the number of nonzero terms in the polynomial, $N_i$ and $a_{Ni}\;(i=1,2,?,K)$ are the exponents and coefficients, respectively. It is given that $1≤K≤10, 0≤N_K <?<N_2<N_1 ≤1000$ .
Output Specification:
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.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 3 3.6 2 6.0 1 1.6

思路||总结

• 设置20001初始值为0的列表，对于两个因子，将其系数的乘积加在列表位置为两个因子的幂的和的位置。
• count一下又多少个非零项，然乎输出。

AC代码

import numpy as npdef main():str_0 = input().strip().split()str_0[0] = int(str_0[0])for i in range(str_0[0]):str_0[i * 2 + 1] = int(str_0[i * 2 + 1])str_0[i * 2 + 2] = float(str_0[i * 2 + 2])str_1 = input().strip().split()str_1[0] = int(str_1[0])for i in range(str_1[0]):str_1[i * 2 + 1] = int(str_1[i * 2 + 1])str_1[i * 2 + 2] = float(str_1[i * 2 + 2])result = np.zeros(2001)count = 0for i in range(str_0[0]):for j in range(str_1[0]):# 对于两个因子，将其系数的乘积加在列表位置为两个因子的幂的和的位置result[str_0[i * 2 + 1] + str_1[j * 2 + 1]] += str_0[i * 2 + 2] * str_1[j * 2 + 2]# 数一下有多少个非0项for i in range(2001):if result[i] != 0:count += 1print(count, end='')for i in range(2001):if result[2000 - i] != 0:print(' %d %.1f' % (2000 - i, result[2000 - i]), end='')main()