1009 Product of Polynomials（25 分）

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 N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N​K​​<⋯<N​2​​<N​1​​≤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

//关键点：多项式乘积之后的范围为0~2000，所以设置double p[2001] = {0};

#include<stdio.h>
#include<iostream>
#include<iomanip>

using namespace std;

int main()
{
    int k1;
    cin >> k1;
    int a[1001] = {0};//存放多项式A的指数
    double b[1001] = {0};//存放多项式A的系数
    for(int i = 0;i< k1;i++)
    {
        cin >> a[i];
        cin >> b[i];
    }

    int k2;
    cin >> k2;
    double p[2001] = {0};//存放乘积的指数对应的系数
    for(int i = 0;i < k2;i++)
    {
        int ex;
        double co;//多项式B的指数，系数
        cin >> ex;
        cin >> co;

        for(int j = 0;j < k1;j++)
        {
            p[ex + a[j]] += co*b[j];
        }
    }

    int count = 0;//结果的项数
    for(int i = 2000;i >= 0;i--)
    {
        if(p[i] != 0)
        {
            count++;
        }
    }
    cout << count;
    for(int i = 2000;i >= 0;i--)
    {
        if(p[i] != 0)
        {
            cout << " " << i << " " << setiosflags(ios::fixed) << setprecision(1) << p[i];
        }
    }
    cout <<endl;

    return 0;
}