# 1009. Product of Polynomials (25)-PAT甲级真题

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.

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

• 输入两个多项式
• k 非0项
• 指数和系数
• 输出他们的乘积，指数降序，小数保留1位
• k 非0项
• 指数和系数

#### 分析

• 输入一个多项式，用arr数组接收
• 输入第二个多项式，输入的时计算该项与第一个多项式各项的乘积ans[j + a] += arr[j] * b;
• 统计非零项的个数cnt
• 输出结果
#include

using namespace std;

int main() {
int n1, n2, a, cnt = 0;
cin >> n1;
double b, arr[1001] = {0.0}, ans[2001] = {0.0};
for (int i = 0; i < n1; i++) {
cin >> a >> b;
arr[a] = b;
}
cin >> n2;
for (int i = 0; i < n2; i++) {
cin >> a >> b;
for (int j = 0; j < 1001; j++)
ans[j + a] += arr[j] * b; //x^(j+a)的系数=(x^j的系数)*(x^a的系数)
}
//统计非零项
for (int i = 2000; i >= 0; i--) {
if (ans[i] != 0)
cnt++;
}
cout << cnt;
for (int i = 2000; i >= 0; i--)
if (ans[i] != 0.0)
printf(" %d %.1f", i, ans[i]);
return 0;
}