PAT A1009 Product of Polynomials (25)

约定不等于承诺〃 2021-09-13 22:22 228阅读 0赞

**题目地址：**[https://www.patest.cn/contests/pat-a-practise/1009][https_www.patest.cn_contests_pat-a-practise_1009]

**题目描述：**

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

**输入格式（Input Specification）：**

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < … < N2 < N1 <=1000.

**输出格式（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.

**输入样例（Sample Input）：**

2 1 2.4 0 3.2  
2 2 1.5 1 0.5

**输出样例（Sample Output）：**

3 3 3.6 2 6.0 1 1.6

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**题意：**

给出两个多项式的系数，求这两个多项式的乘积。

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**解题思路：**

先获得第一个多项式的系数，然后再输入第二个系数时循环与第一个多项式的系数相乘，并将结果加到对应指数的系数上，最后得到要输出的所有非零系数的项。

> **注意点：**
> 
>  *  答案的系数数组要至少开到 2001，因为两个最高幂次为 1000 的多项式相乘，最高幂次可以达到 2000
>  *  没有必要开两个数组存放两个多项式，然后读完系数之后再处理，只需要第二个多项式的系数边读入边处理
>  *  系数是从大到小输出。
>  *  只需要输出非零系数的项。

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C++完整代码如下：

这里写代码片

[https_www.patest.cn_contests_pat-a-practise_1009]: https://www.patest.cn/contests/pat-a-practise/1009