PAT甲级 1009. Product of Polynomials (25)

1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

多项式乘法

#include
#include
#include
#include
#include

using namespace std;

double s1[1050];
double res[2050];
int main()
{
	int k1,k2;
	while(~scanf("%d",&k1))
	{
		memset(s1,0,sizeof(s1));
		memset(res,0,sizeof(res));
		for(int i=1;i<=k1;i++)
		{
			int a;
			double b;
			scanf("%d%lf",&a,&b);
			s1[a]=b;
		}
		scanf("%d",&k2);
		for(int i=1;i<=k2;i++)
		{
			int a;
			double b;
			scanf("%d%lf",&a,&b);
			for(int j=0;j<1001;j++)
			{
					res[j+a]+=s1[j]*b;
			}
		}
		int cnt=0;
		for(int i=0;i<2002;i++)
		{
			if(res[i]!=0) cnt++;
		}
		printf("%d",cnt);
		for(int i=2002;i>=0;i--)
		{
			if(res[i]!=0)
				printf(" %d %.1lf",i,res[i]); 
		}
		putchar(10);
		
	}
	return 0;
}