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

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:

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=0;i<vec.size();i++){
        if(vec[i].expo==po.expo){
            vec[i].coef+=po.coef;
            if(vec[i].coef==0){
                vec.erase(vec.begin()+i);
                /**这边需要判断0多项式,进行erase掉*/
            }
            return;
        }
    }
    if(po.coef!=0) 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=0;i<M;i++){
        cin>>p[i].expo>>p[i].coef;
    }
    cin>>N;
    poly p2[N];
    for(int i=0;i<N;i++){
        cin>>p2[i].expo>>p2[i].coef;
    }
    for(int i=0;i<M;i++){
        for(int j=0;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=0;i<res.size();i++){
        printf(" %d %.1f",res[i].expo,res[i].coef);
    }
    system("pause");
    return 0;
}

需要记住vector进行删除元素,用的是erase(iter*)

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转载自www.cnblogs.com/littlepage/p/11280192.html