PAT Advanced 1009 Product of Polynomials (25 分)（vector删除元素用的是erase）

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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 (,) are the exponents and coefficients, respectively. It is given that 1, 0.

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 <iostream>

#include <vector>

#include <algorithm>

using namespace std;

struct poly{

    int expo;

    double coef;

};

void add(vector<poly>& vec,poly po){

    for(int i=;i<vec.size();i++){

        if(vec[i].expo==po.expo){

            vec[i].coef+=po.coef;

            if(vec[i].coef==){

                vec.erase(vec.begin()+i);

                /**这边需要判断0多项式，进行erase掉*/

            }

            return;

        }

    }

    if(po.coef!=) vec.push_back(po);

}

bool cmp(poly p,poly p2){

    return p.expo>p2.expo;

}

int main(){

    /**

    * 注意点

    * 1.保留一位小数

    */

    int M,N;vector<poly> res;

    cin>>M;

    poly p[M];

    for(int i=;i<M;i++){

        cin>>p[i].expo>>p[i].coef;

    }

    cin>>N;

    poly p2[N];

    for(int i=;i<N;i++){

        cin>>p2[i].expo>>p2[i].coef;

    }

    for(int i=;i<M;i++){

        for(int j=;j<N;j++){

            poly temp;

            temp.expo=p[i].expo+p2[j].expo;

            temp.coef=p[i].coef*p2[j].coef;

            add(res,temp);

        }

    }

    /**这边需要sort一下*/

    sort(res.begin(),res.end(),cmp);

    cout<<res.size();

    for(int i=;i<res.size();i++){

        printf(" %d %.1f",res[i].expo,res[i].coef);

    }

    system("pause");

    return ;

}