1009. Product of Polynomials

最新推荐文章于 2024-03-04 18:04:28 发布

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最新推荐文章于 2024-03-04 18:04:28 发布

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本文详细阐述了如何通过输入两个多项式的系数来计算它们的乘积，并提供了具体实现代码，包括多项式输入、处理和输出步骤。

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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

#include<iostream>
#include<iomanip>
#include<cstring>
using namespace std;
#define max 1000
double arr1[max+1];
double arr2[max+1];
double arr3[2*max+1];
int main()
{
	memset(arr1,0,sizeof(arr1));  
    memset(arr2,0,sizeof(arr2));  
    memset(arr3,0,sizeof(arr3)); 
	int m,n,t;
	double p;
	cin>>m;
	for(int i=0;i<m;++i)
	{
		cin>>t>>p;
		arr1[t]=p;
	}
	cin>>n;
	for(int i=0;i<n;++i)
	{
		cin>>t>>p;
		arr2[t]=p;
	}
	for(int i=0;i<=1000;++i)
	{
		for(int j=0;j<=1000;++j)
			arr3[i+j]+=arr1[i]*arr2[j];//注意使用+=，使用=结果为0.
	}
	int cnt=0;
	for(int i=0;i<=2000;++i)
	{
		if(arr3[i]!=0)
			cnt++;
	}
	cout<<cnt;
	for(int i=2000;i>=0;--i)
	{
		if(arr3[i]!=0.0)
			cout<<" "<<i<<" "<<fixed<<setprecision(1)<<arr3[i];
	}
	cout<<endl;
	return 0;
}