1002 A+B for Polynomials

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本文详细解析了如何通过数组维护和相加两个多项式的算法，输入包含多项式的非零项，输出为相加后的多项式表达式，精确到小数点后一位。

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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:

$K, N_1, a_{N_1}, N_2a_{N_2}, ... , N_Ka_{N_K}$

where $K$ is the number of nonzero terms in the polynomial, $N_i$ and $a_{N_i}$ $(i = 1, 2, \dots, K)$ 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

解题思路：

只需维护一个数组，数组下标表示多项式的幂次，数组值表示系数；先读入 $A$ 的值，之后在读入B时，进行逐个相加，最后输出系数不为 0 的项即可，注意结构体数组初始化时所有系数皆为 0；

#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
	double A[1001] = { 0 };
	int K, exp;
	double coff;
	cin >> K;
	for (int i = 0; i < K; i++) {
		cin >> exp >> coff;
		A[exp] = coff;
	}
	cin >> K;
	for (int i = 0; i < K; i++) {
		cin >> exp >> coff;
		A[exp] += coff;
	}
	int cont = 0;
	for (int i = 0; i < 1001; i++)
		if (A[i] != 0)
			cont++;
	cout << cont;
	for (int i = 1000; i >= 0; i--) {
		if (A[i] != 0) 
			cout << " " << i << " " << fixed << setprecision(1) << A[i];
	}
}