1009 Product of Polynomials

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本文介绍了一种计算两个多项式相乘的算法，并通过C++代码实现了该算法。输入包含两个多项式的非零项系数及指数，输出为这两个多项式相乘后的结果。

1009 Product of Polynomials （25 分)

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 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 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

#include <bits/stdc++.h>
using namespace std;
double N[2005];
int main()
{
	int n;
	vector <int> v1,v3;
	vector <double> v2,v4;
	memset(N,0,sizeof(N));
	cin>>n;
	for(int i=0;i<n;i++)
	{
		int a;
		double b;
		cin>>a>>b;
		v1.push_back(a);
		v2.push_back(b);
	}
	cin>>n;
	for(int i=0;i<n;i++)
	{
		int a;
		double b;
		cin>>a>>b;
		v3.push_back(a);
		v4.push_back(b);
	}
	for(int i=0;i<v1.size();i++)
	{
		for(int j=0;j<v3.size();j++)
		{
			N[v1[i]+v3[j]]+=v2[i]*v4[j];
		}
	}
	int t=0;
	for(int i=0;i<=v1[0]+v3[0];i++)
	{
		if(N[i]!=0)
			t++;
	}
	cout<<t;
	for(int i=v1[0]+v3[0];i>=0;i--)
	{
		if(N[i]!=0)
			printf(" %d %.1f",i,N[i]);
	}
	return 0;
}