PAT 甲级 1009 Product of Polynomials-白红宇的个人博客

发布日期：2021-07-01 03:08:45 浏览次数：2 分类：技术文章

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1009 Product of Polynomials （25 point(s)）

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 3.22 2 1.5 1 0.5

Sample Output:

3 3 3.6 2 6.0 1 1.6

Experiential Summing-up

This question is 's enhanced version. So the difficulty of it is much more than the last, but just a little bit. this question can not be solved by hash array because of complexity. The best method is use a STL Map to record the first polynominals and use another to record the product of two polynominals. Be careful of the item whose second value is equal to zero~

(The purpose of using English to portray my solution is that to exercise the ability of my expression of English and accommodate PAT advanced level's style.We can make progress together by reading and comprehending it. Please forgive my basic grammar's and word's error. Of course, I would appreciated it if you can point out my grammar's and word's error in comment section.( •̀∀•́ ) Furthermore, Big Lao please don't laugh at me because I just a English beginner settle for CET6 _(:з」∠)_ )

Accepted Code

#include 
   
    #include 
    using namespace std;int main(){	int n,k,e;	double coe;	scanf("%d",&k);	map
     
       mp1,mp2;	for(int i=0;i
      
       ::iterator it=mp1.begin();it!=mp1.end();++it)		{			if(mp2.count(it->first+e)==0)				mp2[it->first+e]=it->second*coe;			else				mp2[it->first+e]+=it->second*coe;		}	}	for(map
       
        ::iterator it=mp2.begin();it!=mp2.end();++it)	{		if(it->second==0)			mp2.erase(it->first);	}	printf("%d",mp2.size());	for(map
        
         ::reverse_iterator it=mp2.rbegin();it!=mp2.rend();++it) { printf(" %d %.1f",it->first,it->second); } printf("\n"); return 0;}

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