1009. Product of Polynomials (25)

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本文深入探讨了深度学习和人工智能领域的最新进展，包括机器学习算法、神经网络架构、自然语言处理、计算机视觉等核心主题。文章还涵盖了实际应用案例，展示了这些技术在解决现实世界问题上的潜力。

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

提交代码

题解:　这是两个多项式相乘求解，　数据量不大，　只有1000，　我们可以把两个多项式的积存储到另一个数组当中，　然后统计指数不为0的个数，　最后从大到小依次输出即可，　注意，　指数为0的不输出．　另外如果指数不为0的个数也为0的话，　只打印一个0即可．

#include <cstdio>
#include <cstring>

const int N = 2010;
const double EPS = 1e-8;

void read(double* ary, int n) {
	int idx;
	while (n--) {
		scanf("%d", &idx);
		scanf("%lf", ary + idx);
	}
}

int main() {
	double A[N], B[N], ans[N] = {0.0};

	int n;
	int idx;
	double tmp;
	scanf("%d", &n);
	read(A, n);
	scanf("%d", &n);
	read(B, n);

	for (int i = 0; i <= 1000; ++i)
		for (int j = 0; j <= 1000; ++j) 
			ans[i + j] += A[i] * B[j];

	int cnt = 0;
	for (int i = 0; i < N; ++i)
		if (!(ans[i] >= -EPS && ans[i] <= EPS))
			cnt++;

	printf("%d", cnt);

	for (int i = N - 1; ~i; --i)
		if (!(ans[i] >= -EPS && ans[i] <= EPS))
			printf(" %d %.1f", i, ans[i]);

	return 0;
}