PAT-1009. Product of Polynomials (25)

分析：简单题。相乘时指数相加，系数相乘即可，输出时按指数从高到低的顺序。注意点：多项式相乘后指数最高可达2000。

题目描述：

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 aN1 N2 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

参考代码：

#include





  
#include  
#include  
using namespace std;  
  
#define max 1000  
double input1[max + 1];  
double input2[max + 1];  
double result[2*max + 1];  
  
int main()  
{  
    memset(input1,0,sizeof(input1));  
    memset(input2,0,sizeof(input2));  
    memset(result,0,sizeof(result));  
    int k;  
    int i,j;  
    int e;  
    //int temp=0; //记录最高指数  
    double c;  
    int count = 0; //最终输出的多项式的项数。  
    cin>>k;  
    for(i=0; i>e>>c;  
        input1[e] += c;  
    }  
    cin>>k;  
    for(i=0; i>e>>c;  
        input2[e] += c;  
    }  
  
    for(i=0; i<=1000; i++)  
    {  
        for(j=0; j<=1000; j++)  
        {  
            result[i+j] += input1[i]*input2[j];  
        }  
    }  
    for(i=0; i<=2000; i++) if(result[i] != 0) count++;  
    cout<= 0; i--)  
    {  
        if(result[i] != 0.0) {  
            cout<<" "<