[PAT-甲级]1009.Product of Polynomials

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 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

解题思路：PAT1002. 考察点是两个多项式的加法计算，本题考察的是两个多项式的乘法计算。

• 最高项是1000，所有两个多项式相乘得到的最高项是<=2000的。
• 只需要输出非0项，即系数为0，不需要输出
• 读入第2个多项式的时候，边读入边计算

代码如下：

#include<stdio.h>

// 定义结构体ploy数组保存第一组多项式的输入
struct Poly
{
	int exp;
	double cof;
}poly[1010];

// 保存运算的结果
double ans[2010];

int main()
{
	int n, m, count = 0;
	// 读入第一个多项式
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%d %lf", &poly[i].exp, &poly[i].cof);
	
	// 读入第二个多项式
	scanf("%d", &m);
	for (int i = 0; i < m; i++)
	{
		int exp;
		double cof;

		// 边读入边计算，每读入项都需要跟第一个多项式中每一项相乘
		scanf("%d %lf", &exp, &cof);
		for (int j = 0; j < n; j++)
			ans[exp + poly[j].exp] += (cof*poly[j].cof);
	}

	// 统计非0项的个数
	for (int i = 0; i <= 2000; i++)
	{
		if (ans[i] != 0.0)
			count++;
	}

	// 从高到低打印结果
	printf("%d", count);
	for (int i = 2000; i >= 0; i--)
	{
		if (ans[i] != 0.0)
			printf(" %d %.1f", i, ans[i]);
	}
	
	return 0; 
}