PAT甲级1009 Product of Polynomials

写的很抽象，不应该用map的，应该直接两层for遍历

题目地址：

1009 Product of Polynomials - PAT (Advanced Level) Practice (pintia.cn)https://pintia.cn/problem-sets/994805342720868352/exam/problems/type/7?problemSetProblemId=994805509540921344&page=0
介绍

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​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤NK​<⋯<N2​<N1​≤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.

代码

    #include<bits/stdc++.h>
	using namespace std;
	#define ll long long  
	#define ull unsigned long long
	
	void solve() {
		ll n,m;
        cin>>n;
        map<int,double>ssr;
        double i2;
        int i1;
        for(int i=0;i<n;i++)
        {
            cin>>i1>>i2;
            ssr.insert(pair<int,double>(i1,i2));
        }
        map<int,double>num1;
        cin>>m;
        for(int i=0;i<m;i++)
        {
            cin>>i1>>i2;
            for(auto i=ssr.begin();i!=ssr.end();i++)
            {
                int o1=i->first;
                double o2=i->second;
                if(num1.find(o1+i1)==num1.end())
                num1.insert(pair<int,double>(i1+o1,i2*o2));
                else
                {
                    double key1=num1.find(o1+i1)->second+i2*o2;
                    
                    /*num1.insert(pair<int,double>(i1+o1,key1));*/
                    num1[i1+o1]=key1;
                }
            }
        }
        i1=0;
        int i3=0;
        vector<int> num;
        for(auto i=num1.begin();i!=num1.end();i++)
        {
            int o1=i->first;
            num.push_back(o1);
            if(i->second!=0)
            {
                i3++;
            }
            i1++;
        }
        sort(num.begin(),num.end());
        cout<<i3;
        for(int i=i1-1;i>=0;i--)
        {
            if(num1.find(num[i])->second==0)
            {
                continue;
            }
            cout<<" "<<num[i]<<" ";
            printf("%.1lf",num1.find(num[i])->second);
        }
	}
	signed main() {
	
	    ll t = 1; 
	    // std::cin >> t;
	    while (t--) {
	        solve();
	    }
	}