1009. Product of Polynomials (25)

最新推荐文章于 2024-03-04 18:04:28 发布

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最新推荐文章于 2024-03-04 18:04:28 发布

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本文介绍了一种使用 map 结构实现的多项式乘法算法，该算法能够有效地处理两个多项式的相乘，并输出结果多项式的系数和指数。通过示例输入输出展示了算法的具体应用。

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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 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 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

有人用数组来存储多项式，用数组下标来表示指数，这对于大量的数据是有优势的，此题使用map来映射指数和系数

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
/* 两个多项式的乘积
 *
 */
struct Poly
{
	int ex;
	double co;
};
Poly A[10];
Poly B[10];
map<int, double> C;
int main(void)
{
	int m, n;
	scanf("%d", &m);
	for (int i = 0; i < m; ++i)
		scanf("%d%lf", &A[i].ex, &A[i].co);
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
		scanf("%d%lf", &B[i].ex, &B[i].co);
	int exp;
	for (int i = 0; i < m; ++i)
	{
		for (int j = 0; j < n; ++j)
		{
			exp = A[i].ex + B[j].ex;
			C[exp] += A[i].co * B[j].co;
			if (C[exp] == 0)
				C.erase(exp);
		}
	}
	printf("%d", C.size());

	for (map<int,double>::iterator iter = C.end() ; iter !=  C.begin(); )
	{
		--iter;
		printf(" %d %.1lf", iter -> first, iter -> second);
	}
	printf("\n");
	return 0;