# PAT (Advanced Level) Practice 1009 Product of Polynomials

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
``````

### 题目分析

NK最大是1000，所以乘积中指数最大为2000.

### AC代码

``````#include <bits/stdc++.h>
using namespace std;
struct node
{
int n;
double v;
} a;
double ans = {0};
int main()
{
int i, j, k, cnt = 0;
for (i = 0; i < 2; i++)
{
cin >> k;
for (j = 0; j < k; j++)
cin >> a[i][j].n >> a[i][j].v;
}
for (i = 0; i < 10; i++)
for (j = 0; j < 10; j++)
ans[a[i].n + a[j].n] += a[i].v * a[j].v;
for (i = 0; i <= 2000; i++)
{
if (ans[i])
cnt++;
}
cout << cnt;
for (i = 2000; i >= 0; i--)
{
if (ans[i])
printf(" %d %.1f", i, ans[i]);
}
return 0;
}
``````