009 Product of Polynomials (25 分)
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 aNKwhere 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
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和A1002 异曲同工之处,有点堆排序的意思。
#include<iostream>
using namespace std;
double p[2002];
struct node
{
int zhishu;
double xishu;
}poly[1001];
double ans[2002];
int main()
{
int n,m, count = 0;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> poly[i].zhishu >> poly[i].xishu;
}
cin >> m;
for (int i = 0; i < m; i++)
{
int zhishu; double xishu;
cin >> zhishu >> xishu;
for (int j = 0; j < n; j++)
ans[zhishu + poly[j].zhishu] += (xishu * poly[j].xishu);
}
for (int i = 0; i < 2002; i++)
{
if(ans[i] != 0)
count++;
}
cout << count;
for (int i = 2001; i >=0; i--)
{
if (ans[i] != 0.0)
printf(" %d %.1f", i, ans[i]);
}
}