PAT 1009 Product of Polynomials

1009 Product of Polynomials (25分)

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

 

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

 

 

思路

这道题是一个多项式的乘法，考察map的使用和如何定义按照指定规则排序的map，

利用map进行遍历即可，也可以使用数组的方法，注意系数为0的情况要被删除！

**这题坑在测试点1,有系数为0的项。就是多项式的每一项乘另外多项式的每一项然后相加,然后合并同类项之后为0 ！！！

 

#include <iostream>
#include <stdio.h> 
#include <map>
#include <algorithm>
#include <iterator>


using namespace std;

double a[30];
double b[30];
map<int, double, greater<int> > mp;

int main()
{
    int ka, kb;
    cin >> ka;
    ka = 2*ka;
    for(int i = 1; i <= ka; i++)
    {
        cin >> a[i];
    }
    
    cin >> kb;
    kb = 2*kb;
    for(int i = 1; i <= kb; i++)
    {
        cin >> b[i];
    }
    
        
    for(int i = 1; i <= ka; i+=2)
    {
        for(int j = 1; j <= kb; j+=2)
        {
            mp[a[i]+b[j]] += a[i+1]*b[j+1];
            if(mp[a[i]+b[j]] == 0)    // 合并后系数有可能为0，此时要删去
                mp.erase(a[i]+b[j]);
        }
    }
    
    int Size = mp.size();
    printf("%d", Size);
    for(auto iter = mp.begin(); iter != mp.end(); ++iter)
    {
        printf(" %d %.1f", iter->first, iter->second);
    }


    return 0;
}