PATA 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

#include <cstdio>
const int maxn = 2001;
int main()
{
	double coef[maxn]={0};
	double ans[maxn]={0};  //原来直接用coef[]存储计算后的系数，导致出现问题，应另用一个ans保存 。
	int k1,k2,i,j,ex,k3=0;
	double co;
	scanf("%d",&k1);
	for(i = 0;i < k1; i++)
	{
		scanf("%d%lf",&ex,&co);
		coef[ex] = co;
	}
	scanf("%d",&k2);
	for(i = 0;i < k2; i++)
	{
		scanf("%d%lf",&ex,&co);
		for(j = 0;j <1001;j++)
		{
			ans[ex+j] += co*coef[j];
		}
	}
	for(i = 0;i < maxn; i++)
	{
		if(ans[i] != 0.0) k3++;
	}
	printf("%d",k3);

	for(i = maxn-1;i >= 0; i--)
	{
		if(ans[i]!=0.0){
			printf(" %d %.1lf",i,ans[i]);
		}
	}
	return 0;
}