甲级-1009 Product of Polynomials (25 分)

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 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.

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

解题思路:

两层循环,分别将两个序列的每个元素的系数相乘、指数相加,最后再将具有相同指数的系数相加合并即可。
注意在此处只需要将系数真正为0的元素去除,而不需要将系数<0.05的元素去除,虽然后者在保留一位小数的情况下显示出来为0.0。

代码:

编译器:C++(g++)

#include <iostream>
#include <vector>
#include <map>
using namespace std;

int main()
{
    int k;
    cin>>k;
    map<int,double> fi;
    for(int i=0;i!=k;++i)
    {
        int t1;
        double t2;
        cin>>t1>>t2;
        fi[t1]=t2;
    }
    cin>>k;
    map<int,double> si;
    for(int i=0;i!=k;++i)
    {
        int t1;
        double t2;
        cin>>t1>>t2;
        si[t1]=t2;
    }
    map<int,double> result;
    for(auto i=fi.begin();i!=fi.end();++i)
    {
        for(auto j=si.begin();j!=si.end();++j)
        {
            result[i->first+j->first]+=i->second*j->second;
        }
    }
    //删除系数为0的项
    for(auto i=result.begin();i!=result.end();)
    {
        if(i->second<=1e-15&&i->second>=-1e-15)
            i=result.erase(i);
        else
            ++i;
    }
    cout<<result.size();
    cout.flags(ios::fixed);
    cout.precision(1);
    for(auto i=result.rbegin();i!=result.rend();++i)
    {
        cout<<" "<<i->first<<" "<<i->second;
    }
    cout<<endl;
    return 0;
}
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