A1009. Product of Polynomials (25)

最新推荐文章于 2022-02-25 22:40:25 发布

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最新推荐文章于 2022-02-25 22:40:25 发布

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本文介绍了一种计算两个多项式乘积的方法，并提供了两种不同的实现方案。通过使用数组和结构体来存储多项式的幂次和系数，实现了高效的计算。

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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

题目大意：

计算两个多项式的乘积。

思路：

（1）用数组下标存放幂次，下标对应的元素存放系数。double型数组。最高幂次可达到2000，数组应大于2000。

因有些幂次最后可能为0，非零项应最后统计。本题不需考虑多项式相乘的顺序；

（2）可加入结构体，存入第一组数据，在读入第二组数据的同时直接处理。

（1）代码如下：

#include <cstdio>

const int max_n = 2010;
int main()
{
	int k, exp, count = 0;
	double coe;	
	double a[max_n] = {}, b[max_n] = {}, c[max_n] = {};	 
	
	scanf ("%d", &k);
	while (k--)
	{
		scanf ("%d %lf", &exp, &coe);
		a[exp] = coe;							//数组下标为幂次，其中元素为该幂次的系数； 
	}
	
	scanf ("%d", &k);
	while (k--)
	{
		scanf ("%d %lf", &exp, &coe);
		b[exp] = coe;	
	}
	
	for (int i = max_n - 1; i >= 0 ; i--)			//a[max_n]处存放空字符'\0'，非0，不应加入考虑； 否则会影响对非零项的统计；
	{
		if (a[i] != 0)						//系数为零项，相乘无意义，下同。 
		{
			for (int j = max_n - 1; j >= 0; j--)
			{
				if (b[j] != 0)
					c[i + j] += a[i] * b[j];	//相乘后的幂次即为两幂次的和； 
			}
		}
	}

	for (int i = 0; i < max_n; i++)		
	{
		if (c[i] != 0)		count++;  			//累计非零项系数的项数； 
	}

	printf ("%d", count);
	for (int i = max_n - 1; i >= 0; i--)
	{
		if (c[i] != 0)
		{
			printf (" %d %.1f", i, c[i]);         
		}
	}
	
	return 0;
}

（2）代码如下：

#include <cstdio>

struct poly
{
	int exp;
	double cof;
} poly[1001]; // 存入第一个多项式 

double ans[2010] = {};
int main()
{
	int n, m, number = 0;
	scanf ("%d", &n);
	for (int i = 0; i < n; i++)
	{
		scanf ("%d %lf", &poly[i].exp, &poly[i].cof);
	}
	
	scanf ("%d", &m);                    //两次的多项式项数要分开记录；用于记录存入数据的结构体的项数；
	for (int i = 0; i < m; i++)
	{
		int exp;
		double cof;
		scanf ("%d %lf", &exp, &cof);
		for (int j = 0; j < n; j++)            //存入数据与第一组多项式所有幂次相乘；
		{
			ans[poly[j].exp + exp] += poly[j].cof * cof;
		}
	}
	
	for (int i = 0; i <= 2000; i++)
	{
		if (ans[i] != 0.0)	number++;
	}

	printf ("%d", number);
	for (int i = 2000; i >= 0 ; i--)                                //依据题意从大到小输出
	{
		if (ans[i] != 0.0)
		{
			printf (" %d %.1f", i, ans[i]);        
		}
	}
	
	return 0;
}