PAT A1009. Product of Polynomials (25)

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.

2 1 2.4 0 3.22 2 1.5 1 0.5Sample Output
3 3 3.6 2 6.0 1 1.6
#include <cstdio>#include <algorithm>#define Max 2010using namespace std;int main(){	double a[Max],b[Max],f,c[Max];	for(int i=0;i<Max;i++)	{		a[i]=b[i]=c[i]=0.0;	}	int n,m,k=0,p=0;	scanf("%d",&n);	for(int i=0;i<n;i++)	{		scanf("%d%lf",&m,&f);		a[m]+=f;	}	scanf("%d",&n);	for(int i=0;i<n;i++)	{		scanf("%d%lf",&m,&f);		b[m]+=f;	}	for(int i=0;i<Max;i++)	{		if(a[i]!=0.0)		{			for(int j=0;j<Max;j++)			{				if(b[j]!=0.0)				{					c[i+j]+=a[i]*b[j];					if(i+j>p)					{						p=i+j;					}				}			}		}			}    for(int i=0;i<Max;i++)	{		if(c[i]!=0.0)			k++;	}    if(k==0) printf("0 0 0/n");	else {		printf("%d ",k);		for(int i=p;i>=0;i--)		{			if(c[i]!=0.0)			{				printf("%d %.1f",i,c[i]);				k--;				if(k>0)					printf(" ");				else printf("/n");			}		}	}	system("pause");	return 0;}