PAT A 1009. Product of Polynomials (25)

原題

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
      

求兩個多項式的乘積，直接模拟。

跳過無效項即可，題目沒說清，四舍五入後為0.0的項不顯示。

代碼：

#include <iostream>
using namespace std;

int main()
{
	int k1,k2,k=0;		//多項式1、2、乘積有效項
	double poly1[1001];	//多項式1，标号對應次數
	double poly2[1001];	//多項式2
	double poly[2001];	//乘積
	
	int i,j,exp;
	double coe;
	for(i=0;i<1001;i++)	//初始化
	{
		poly1[i]=0.0;
		poly2[i]=0.0;
	}
	for(i=0;i<2001;i++)
		poly[i]=0.0;
	
	cin>>k1;
	for(i=0;i<k1;i++)	//輸入
	{
		cin>>exp>>coe;
		poly1[exp]=coe;
	}
	cin>>k2;
	for(i=0;i<k2;i++)
	{
		cin>>exp>>coe;
		poly2[exp]=coe;
	}
	for(i=0;i<1001;i++)	//相乘
		for(j=0;j<1001;j++)
			poly[i+j]+=poly1[i]*poly2[j];

	i=2000;
	while(i>=0&&poly[i]<0.05&&poly[i]>-0.05)
		i--;
	if(i<0)
		cout<<0;
	else
	{
		j=i;
		while(j>=0)	//統計有效項
		{
			if(poly[j]>=0.05||poly[j]<=-0.05)
				k++;
			j--;
		}
		cout<<k;
		cout<<fixed;
		cout.precision(1);
		while(i>=0)	//輸出
		{
			if(poly[i]>=0.05||poly[i]<=-0.05)
				cout<<" "<<i<<" "<<poly[i];
			i--;
		}
	}

	return 0;
}