【PAT 】PAT A1002 A+B for Polynomials 多项式相加【多项式加法】

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​N​1​​​​ N​2​​ a​N​2​​​​ … 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,⋯,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 sum 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 to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 2 1.5 1 2.9 0 3.2

题目大意

将两个多项式相加，输出不为0的项数， 以及各项的幂和底数。

思路

根据幂的取值范围，我们开一个长度为1001的double数组来存储各位的底数。使用min和max两个变量来记录最小幂和最大幂，优化性能（其实遍历1001项也不是很慢）。更多细节看代码注释。

代码

#define err 1e-7
#include <math.h>
#include <stdio.h>
int main() {
   
   
    double a[1001]