PAT (Advanced Level)1009. Product of Polynomials (25) 水

题目链接

1009. Product of Polynomials (25)

Time limit:400 ms Memory limit:65536 kB

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Problem Descrpition

This time, you are supposed to find A*B where A and B are two polynomials.

Input

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
K N1 a_N1 N2 a_N2 … NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < … < N2 < N1 <=1000.

Output

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

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题意

给出A，B两个多项式，求A*B

解题思路

本来以为做了A+B之后A×B应该没问题的，，结果还是踩坑，特此记录！

exponents ：指数
coefficients ：系数

另外我是用map来存结果，所以每次更新后，若值为0，则需要将这个元素删除

Code

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <map>
#include <string>

using namespace std;

map<int,double> arr1,arr2,res;

int main()
{
    int n,a;
    double x;
    cin>>n;
    while(n--)
    {
        cin>>a>>x;
        arr1[a]=x;
    }
    cin>>n;
    while(n--)
    {
        cin>>a>>x;
        arr2[a]=x;
    }
    map<int,double>::iterator i,j;
    for(i=arr1.begin();i!=arr1.end();i++)
    {
        for(j=arr2.begin();j!=arr2.end();j++)
        {
            int key=i->first+j->first;
            double val=i->second*j->second;
            if(!res[key])
                res[key]=val;
            else
                res[key]+=val;
            if(res[key]==0)
            {
                res.erase(key);
            }
        }
    }
    cout<<res.size();
    for(map<int,double>::reverse_iterator it=res.rbegin();it!=res.rend();it++)
    {
        cout<<' '<<it->first;
        printf(" %.1f",it->second);
    }
    return 0;
}