1009 Product of Polynomials (25 分)
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 aN1N2 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
#include <cstdio>
#include <cstring>
#include <cmath>
double A[1001] = { 0 }, B[1001] = { 0 };//下标存放指数,内容为系数
double result[2001] = { 0 };//存放乘积的指数系数,指数最大为1000+1000
int main()
{
int n = 0;
int exp = 0;
double coe = 0;
int cnt = 0;
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%d %lf", &exp, &coe);
A[exp] = coe;
}
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
scanf("%d %lf", &exp, &coe);
B[exp] = coe;
}
for (int i = 0; i <= 1000; i++)
{
for (int j = 0; j <= 1000; j++)
{
if (A[i] && B[j])
{
result[i + j] += A[i] * B[j];//乘法规则,指数相加,系数相乘;注意为累加
}
}
}
for (int i = 0; i < 2001; i++) if (result[i]) cnt++;//计算项数
printf("%d", cnt);
if(!cnt) printf(" 0 0.0");//最终结果为0;不过貌似测试用例没测试这个的
for (int i = 2000; i >= 0; i--)
{
if (result[i])
printf(" %d %.1lf", i, result[i]);//输出即可了
}
return 0;
}