C++ Pat甲级1009 Product of Polynomials （25 分）

1009 Product of Polynomials （25 分）多项式的积

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

 输入两行：项数N  每项的指数Ni,系数aNi

输出 ：两多项式乘积 ，项数N  每项的指数Ni,系数aNi

#include<cmath>
#include<cstdio>
#include<vector>
#include<iostream>
using namespace std;
//pair的两个属性值  first  和second 
#define fi first
#define se second
int main()
{
    int n,k;
    double m;
    double p[2000+5] = {0};     //乘积的最大项的次数为2000
    vector<pair<double,int> > a;//相当于在容器里存了一个struct类型 
    vector<pair<double,int> > b;
    //输入两个多项式 
	cin >> n;
    while (n--)
    {
        cin >> k >> m;
        a.push_back(make_pair(m,k));
    }
    cin >> n;
    while (n--)
    {
        cin >> k >> m;
        b.push_back(make_pair(m,k));
    }
    // 每项分别相乘，两重循环 
    for (int i = 0 ; i < a.size() ; i++)
    {
        for (int j = 0 ; j < b.size() ; j++)
        {
            p[a[i].se + b[j].se] += a[i].fi * b[j].fi;//p[指数]中存的是对应指数的系数 
        }
    }
    //将a的内容删除，将p中系数不为0的项存入a 
    a.clear();
    
    for (int i = 2000 ; i >= 0 ; i--)
    {
        if (fabs(p[i]) >= 0.05)
        {
            a.push_back(make_pair(p[i],i));
        }
    }
//输出结果
    cout << a.size();
    for (int i = 0 ; i < a.size() ; i++)
        printf (" %d %.1lf",a[i].se,a[i].fi);
    cout << endl;
    return 0;
}

C++中pair的使用方法 https://blog.csdn.net/oceanlight/article/details/7890537

pair 和 make_pair 的使用 https://blog.csdn.net/hiwoshixiaoyu/article/details/53894162

多项式加法：

	// 每项分别相加，两重循环
	for (int i = 0 ; i < a.size() ; i++) {
		for (int j = 0 ; j < b.size() ; j++) {
			if(a[i].se == b[j].se) {
				p[a[i].se + b[j].se] = a[i].fi + b[j].fi;//p[指数]中存的是对应指数的系数
			}else if(a[i].se != 0){
				p[a[i].se] = a[i].fi;
			}else if(b[j].se != 0){
				p[b[j].se] = b[j].fi;
			}

		}
	}