【PAT】1009. Product of Polynomials (25)

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1009

分析：简单题。相乘时指数相加，系数相乘即可，输出时按指数从高到低的顺序。注意点：多项式相乘后指数最高可达2000。

题目描述：

 

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

 

参考代码：

 

#include<iostream>

#include<iomanip>

#include<string.h>

using namespace std;



#define max 1000

double input1[max + 1];

double input2[max + 1];

double result[2*max + 1];



int main()

{

	memset(input1,0,sizeof(input1));

	memset(input2,0,sizeof(input2));

	memset(result,0,sizeof(result));

	int k;

	int i,j;

	int e;

	//int temp=0; //记录最高指数

	double c;

	int count = 0; //最终输出的多项式的项数。

	cin>>k;

	for(i=0; i<k; i++)

	{

		cin>>e>>c;

		input1[e] += c;

	}

	cin>>k;

	for(i=0; i<k; i++)

	{

		cin>>e>>c;

		input2[e] += c;

	}



	for(i=0; i<=1000; i++)

	{

		for(j=0; j<=1000; j++)

		{

			result[i+j] += input1[i]*input2[j];

		}

	}

	for(i=0; i<=2000; i++) if(result[i] != 0) count++;

	cout<<count;

	for(i=2000; i >= 0; i--)

	{

		if(result[i] != 0.0) {

			cout<<" "<<i;

			cout<<fixed<<setprecision(1);

			cout<<" "<<result[i];

		}

	}

	cout<<endl;

	return 0;

}