PAT 1009. Product of Polynomials

1009. Product of Polynomials (25)

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

分析

我是用map来做这题的，感觉没什么可说的key为次数，value为系数，map按key的非递减的顺序自动排序，输出时利用rbegin,rend,来输出就行了←_←

代码如下

#include<iostream>
#include<map>
using namespace std;
int main(){
	map<int,double> p1,p2,p;
	int N,exp,cnt=0;
	double coe;
	cin>>N;
	for(int i=0;i<N;i++){
		cin>>exp>>coe;
		p1[exp]=coe;
	}
	cin>>N;
	for(int i=0;i<N;i++){
		cin>>exp>>coe;
		p2[exp]=coe;
	}
	for(auto a=p1.begin();a!=p1.end();a++){
		for(auto b=p2.begin();b!=p2.end();b++){
			p[a->first+b->first]+=a->second*b->second;
		}
	}
	for(auto b=p.rbegin();b!=p.rend();b++)
	if(b->second!=0) cnt++;
	cout<<cnt;
	for(auto b=p.rbegin();b!=p.rend();b++)
	if(b->second!=0) printf(" %d %.1f",b->first,b->second);
}