1009 Product of Polynomials （25 分）_product of polynomials 提交之后一直是部分错误-优快云博客

本文链接：https://blog.youkuaiyun.com/TOBEALISTENNER/article/details/86668256

本文详细介绍了一种解决多项式乘法问题的高效算法。通过输入两个多项式的非零项，算法能准确计算并输出其乘积，结果精确到小数点后一位。示例输入与输出展示了算法的具体应用，代码实现部分使用C++语言，涵盖了输入处理、乘法运算及输出格式的完整流程。

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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:

KKK N1N_1N1 aN1a_{N_1}aN1 N2N_2N2 aN2a_{N_2}aN2 ... NKN_KNK aNKa_{N_K}aNK

where KKK is the number of nonzero terms in the polynomial, NiN_iNi and aNia_{N_i}aNi (i=1,2,⋯,Ki=1, 2, \cdots , Ki=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤101\le K \le 101≤K≤10, 0≤NK<⋯<N2<N1≤10000 \le N_K < \cdots < N_2 < N_1 \le 10000≤NK<⋯<N2<N1≤1000.

Output Specification:

For each test case you should output the product of AAA and BBB 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

注意事项：系数相乘后最大可以为2000

代码如下：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int K[2], N[2][1001];
	float a[2][1001];
	memset(a, 0, sizeof(a));
	for (int p = 0; p < 2; p++)
	{
		cin >> K[p];
		for (int q = 0; q < K[p]; q++)
			cin >> N[p][q] >> a[p][q];
	}

	float resN[2001] = { 0 };
	for (int i = 0; i < K[0]; i++)
	{
		for (int j = 0; j < K[1]; j++)
		{
			resN[N[0][i] + N[1][j]] += a[0][i] * a[1][j];
		}
	}

	int sum = 0;
	for (int k = 0; k <= 2000; k++)
	{
		if (resN[k])sum++;
	}
	cout << sum;
	for (int l = 2000; l >= 0; l--)
	{
		if (resN[l])
			printf(" %d %.1f", l, resN[l]);
	}
	return 0;
}