PAT甲级1009 Product of Polynomials（数组循环）

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本文详细介绍了一种解决多项式相乘问题的高效算法。通过输入两个多项式的非零项系数和指数，算法能准确计算并输出它们的乘积，保留一位小数精度。文章通过具体示例解释了算法的工作原理，并提供了完整的C++代码实现。

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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

简单的多项式相乘然后输出非零项

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int a,b;
	cin>>a;
	double dat[1111];
	double datt[1111];
	double ans[2111];
	memset(dat,0,sizeof(dat));
	memset(datt,0,sizeof(datt));
	memset(ans,0,sizeof(ans));
	double x;
	int s;
	for(int i=0;i<a;i++)
	{
		cin>>s>>x;
		dat[s]=x;
	} 
	cin>>b;
	for(int i=0;i<b;i++)
	{
		cin>>s>>x;
		datt[s]=x;
	}
	for(int i=0;i<=1000;i++)
	{
		for(int j=0;j<=1000;j++)
		{
			ans[i+j]+=dat[i]*datt[j];
		}
	}
	int js=0; 
	for(int i=0;i<=2000;i++)
	if(ans[i]!=0)js++;
	cout<<js;
	for(int i=2000;i>=0;i--)
	{
		if(ans[i]!=0)
		{
			cout<<" "<<i<<" ";
			printf("%.1lf",ans[i]);
		}
	}
	cout<<endl;
	return 0;
}