点击链接PAT甲级-AC全解汇总

题目：
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​​ <⋯​2​​ ​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
题意：
跟PAT甲级1002的输入一样，不过1002是求和，这是求两个多项式的乘积

注意： 乘积的话指数最高可能有2000，数组要开到2000+才行

我的代码：

#include
using namespace std;

int main()
{
    vector<pair<int,double> >exp_ploy;
    int N;
    cin>>N;
    for(int i=0;i<N;i++)
    {
        int exp;
        double ploy;
        cin>>exp>>ploy;
        exp_ploy.push_back(pair<int,double>(exp,ploy));
    }

    double res[2020]={0};
    cin>>N;
    for(int i=0;i<N;i++)
    {
        int exp;
        double ploy;
        cin>>exp>>ploy;
        for(auto it:exp_ploy)
        {
            int t_exp=it.first+exp;
            double t_ploy=it.second*ploy;
            res[t_exp]+=t_ploy;
        }
    }

    int cnt=0;
    for(int i=2000;i>=0;i--)
    {
        if(res[i]!=0)
        {
            cnt++;
        }
    }
    cout<<cnt;
    for(int i=2000;i>=0;i--)
    {
        if(res[i]!=0)
        {
            printf(" %d %.1f",i,res[i]);
        }
    }
    return 0;
}

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