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

     一道简单题，只要注意范围还有当系数为0的时候不输出就好了，数据范围2000，时间限制400ms,然后我就决定使用最蠢的方法。用一个double数组表示多项式，下标为幂指数，内容为系数。相乘的时候逐项相乘，比如第i项与第j项相乘,ans[i+j] += a[i]*b[j];简单愉快。最后扫一遍输出就好了。

# include 
# include 

const int size = 2005;
const double eps = 1e-5; 

double ans[size];
double elem[2][size];

int main()
{
    int i,j,k,n;
    double an;
    for (i=0;ieps)
		      k++;
	printf("%d",k);
	for (i=size-1;i>-1;i--)
	    if (fabs(ans[i])>eps)
		    printf(" %d %.1lf",i,ans[i]);
	printf("\n");        
    return 0;
}