A1002 A+B for Polynomials[简单模拟--多项式相加]

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

Input

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.

Output

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

大意：
程序输入为两行：均为一个多项式，按 K N1 An1 N2 An2…Nk Ank，K代表的是多项式的非零项数，范围闭区间是[1,10]，N1到Nk的范围区间是 1<= Nk <= …<= N1 <= 1000;

Nk是指数，Ank是系数，遇到相同的指数，系数进行累加，从而合并成一个多项式。

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这题自己的写法复杂而且花了很久时间没找出问题，昨晚没睡好有点脑袋发晕，后面看了看最优做法很清晰明了，直接学习了算了，不想整了

#include<iostream>
using namespace std;
double p[1001];
int main()
{
	int k, n, count = 0;
	double a;
	cin >> k;
	for (int i = 0; i < k; i++)
	{
		cin >> n >> a;
		p[n] += a;
	}
	cin >> k;
	for (int i = 0; i < k; i++)
	{
		cin >> n >> a;
		p[n] += a;
	}
	for (int i = 0; i < 1001; i++)
	{
		if (p[i] != 0)
			count++;
	}
	cout << count;
	for (int i = 1000; i >= 0; i--)
		if (p[i] != 0)printf(" %d %.1f",i,p[i]);
	return 0;
}