1009. Product of Polynomials (25)——PAT (Advanced Level) Practise

题目信息：

1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 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 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.

Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

代码如下：

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

int main()
{
	map<int, float> m1, m2, m3;
	int k, n;
	float a;
	cin >> k;
	for (int i = 0; i < k; i++)
	{
		cin >> n >> a;
		m1[n] = a;
	}
	cin >> k;
	for (int i = 0; i < k; i++)
	{
		cin >> n >> a;
		m2[n] = a;
	}
	for (map<int, float>::iterator it1 = m1.begin(); it1 != m1.end(); it1++)
	{
		for (map<int, float>::iterator it2 = m2.begin(); it2 != m2.end(); it2++)
		{
			m3[it1->first + it2->first] += it1->second * it2->second;
		}
	}
	cout.setf(ios_base::fixed, ios_base::floatfield);
	cout.precision(1);
	for (map<int, float>::iterator it = m3.begin(); it != m3.end(); it++)
	{
		if (it->second == 0.0f)
			m3.erase(it);
	}
	cout << m3.size();
	for (map<int, float>::reverse_iterator it = m3.rbegin(); it != m3.rend(); it++)
	{
			cout <<" "<< it->first << " " << it->second;
	}
	cout << endl;
	return 0;
}