1002. A+B for Polynomials (25)

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本文介绍了一种解决多项式加法问题的算法实现方法，输入包含两个多项式的系数及指数，输出为这两个多项式的和。算法使用了数组来存储多项式的各项，并通过逐项相加实现了多项式的加法。

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

两个多项式的加法问题，注意，题目给的输入中最多只有一千项。。。

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
	double v[1001] = {0.0};
	int n1,n2;
	int cnt = 0;
	int index;
	double coe;
	cin >> n1;
	for(int i = 0;i < n1;i++)
	{
		cin >> index >> coe;
		v[index] = coe;
	}
	cin >> n2;
	for(int i = 0;i < n2;i++)
	{
		cin >> index >> coe;
		v[index] += coe;
	}
	for(int i = 0;i < 1001;i++)
		if(fabs(v[i]) > 1e-6)
			cnt++;
	cout << cnt;
	for(int i = 1000;i >= 0;i--)
		if(fabs(v[i]) > 1e-6)
			printf(" %d %.1lf",i,v[i]);
	system("pause");
	return 0;
}