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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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

题意：简单的多项式乘法

思路：一个个乘就好了，这里我并不能完全通过，在牛客网上的测试数据中我发现了自己的问题，就是在多项式的系数相乘过程中，丢失精度，导致最后的一个数据出现0.1的误差，由于水平有限，暂时不知改如何解决

下面是部分AC代码

#include <stdio.h>

struct Node {
	int exp;
	double coe;
} Poly1[1005], Poly2[1005];
double Poly3[2010];
int n, m;

int main()
{
	//初始化 
	for(int i=0;i<2010;i++)
		Poly3[i] = -1;
	//输入 
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%d %lf", &Poly1[i].exp, &Poly1[i].coe);
	scanf("%d", &m);
	for(int i=0;i<m;i++)
		scanf("%d %lf", &Poly2[i].exp, &Poly2[i].coe);
	int count = 0;
	//计算 
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			int temp_exp = Poly1[i].exp + Poly2[j].exp;
			double temp_coe = Poly1[i].coe * Poly2[j].coe;
			if (Poly3[temp_exp] == -1)
			{
				count++;
				Poly3[temp_exp] = temp_coe;
			}
			else
			{
				Poly3[temp_exp] += temp_coe;
			}
		}
	}
	//输出 
	printf("%d", count);
	for (int i = 2009; i >= 0; i--)
	{
		if (Poly3[i] != -1)
		{
			printf(" %d %.1f", i, Poly3[i]);
		}
	
	}
	return 0;
}