题目

This time, you are supposed to find $A\times B$ 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:

$K$ $N_1$ $a_{N_1}$ $N_2$ $a_{N_2}$ … $N_K$ $a_{N_K}$

where $K$ is the number of nonzero terms in the polynomial, $N_i$ and $a_{N_i}$ ( $i=1, 2, \cdots , K$ ) are the exponents and coefficients, respectively. It is given that $1\le K \le 10$ , $0 \le N_K < \cdots < N_2 < N_1 \le 1000$ .

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:

1
2

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

1

3 3 3.6 2 6.0 1 1.6

思路

为了思路简单，就直接开大数组遍历了，没有什么难点

代码

Github最新代码，欢迎交流

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#include <stdio.h>
int main()
{
	int N, exp, count = 0;
	float coef, A[1001] = {0}, B[1001] = {0}, MUL[2001] = {0};

	scanf("%d", &N);
	while (N--) {
		scanf("%d %f", &exp, &coef);
		A[exp] = coef;
	}
	scanf("%d", &N);
	while (N--) {
		scanf("%d %f", &exp, &coef);
		B[exp] = coef;
	}

	for (int i = 0; i < 1001; i++)
		for (int j = 0; j < 1001; j++)
			MUL[i + j] += A[i] * B[j];

	for (int i = 0; i < 2001; i++)
		if (MUL[i]) count++;

	printf("%d", count);
	for (int i = 2000; i >= 0; i--)
		if (MUL[i])
			printf(" %d %.1f", i, MUL[i]);

	return 0;
}