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 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (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

翻译:
和1002类似,只不过这里是多项式相乘,意思就是:如
1(2.4) 0(3.2) 2(1.5) 1(0.5),将指数相加,系数相乘
2+1(2.4*1.5)=>3(3.6)
0+2(3.2*1.5)=>2(4.8)
1+1(2.4*0.5)=>2(4.8+1.2)=>2(6.0)
0+1(3.2*0.5)=>1(1.6)
输出:
3 3 3.6 2 6.0 1 1.6
思路:
第一遍读取后直接存入arr,第二遍每读取一对值,就遍历一遍arr[],如果为arr[]非0,进行运算,并相加,其中第二遍的值保存在ans中,其中
ans的长度设为2001(0~1000+1000)


#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
using namespace std;
int main() 
{
	int n, m, a, cnt = 0;
	scanf("%d", &n);
	float b, arr[1001] = {0}, ans[2001] = {0};
	for (int i = 0; i < n; i++) 
	{
		scanf("%d %f", &a, &b);
		arr[a] = b;
	}
	scanf("%d", &m);
	for (int i = 0; i < m; i++) 
	{
		scanf("%d %f", &a, &b);
		for (int j = 0; j < 1001; j++) 
		{ 
				if (arr[j] != 0) 
				ans[j + a] += arr[j] * b; 
		} 
	} 
	for (int i = 2000; i >= 0; i--)
		if (ans[i] != 0.0) cnt++;
	printf("%d", cnt);
	for (int i = 2000; i >= 0; i--)
	{
		if (ans[i] != 0.0)
			printf(" %d %.1f", i, ans[i]);
	}
	return 0;
}