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

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

1009. Product of Polynomials (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue


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

解读

从多项式乘法的法则出发,就是对每一项相乘之后求和。然后再将对应的结果放入存储的位置中去。

实现方法和思维与加法类似。用顺序表来存放最终结果,病输出系数不为零的项。

完整代码

#include<iostream>  
#include<iomanip> 
#include<vector>  
#include<algorithm>  
using namespace std;

typedef struct Node
{
	double sum;
	int e;
}Node;
#define MAX 3000  
int main()
{
	int n1, n2;
	cin >> n1;

	std::vector<Node> a(n1);
	for (int i = 0; i < n1; ++i)
	{
		cin >> a[i].e;
		cin >> a[i].sum;
	}
			
	cin >> n2;
	std::vector<Node> b(n2);
	for (int i = 0; i < n2; ++i)
	{
		cin >> b[i].e;
		cin >> b[i].sum;
	}
	//product  
	std::vector<double> p;
	p.assign(MAX + 1, 0.0);
	for (int i = 0; i < n1; ++i)
	{
		for (int j = 0; j < n2; ++j)
		{
			Node tmp;
			tmp.e = a[i].e + b[j].e;
			tmp.sum = a[i].sum*b[j].sum;
			p[tmp.e] += tmp.sum;
		}
	}
	//output  
	int cnt = 0;
	for (int i = MAX; i >= 0; --i)
	{
		if (std::fabs(p[i]) > 1e-6)
			cnt++;
	}
	printf("%d", cnt);
	for (int i = MAX; i >= 0; --i)
	{
		if (std::fabs(p[i]) > 1e-6)
			cout << fixed << setprecision(1) << " " << i << " " << p[i];
	}
	printf("\n");
	
	return 0;
}

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