浙大 PAT Advanced level 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

多项式的乘法，用链表做比较复杂，用map容器可以做的非常简单，记得最后剔除系数为0的项即可

#include <iostream>
#include <map>
#include <iomanip>
using namespace std;

int main()
{
map<int, float> poly_fir;
map<int, float> poly_sec;
map<int, float> result;

int n;
int exp;
float coe;
int count = 0;

cin >> n;
while(n--)
{
cin >> exp >> coe;
poly_fir.insert(make_pair(exp, coe));
}
cin >> n;
while(n--)
{
cin >> exp >> coe;
poly_sec.insert(make_pair(exp, coe));
}

for (map<int, float>::const_iterator it_fir = poly_fir.begin();
it_fir != poly_fir.end(); ++it_fir)
{
for (map<int, float>::const_iterator it_sec = poly_sec.begin();
it_sec != poly_sec.end(); ++it_sec)
{
result[it_fir->first + it_sec->first] += it_fir->second * it_sec->second;
}
}

for (map<int, float>::iterator iter = result.begin(); iter != result.end(); )
{
if (0 == iter->second)
{
result.erase(iter++);
}
else
{
++iter;
++count;
}
}
cout << count;

for (map<int, float>::reverse_iterator iter = result.rbegin(); iter != result.rend(); ++iter)
{
cout << ' ' << iter->first << ' ';
cout << fixed << setprecision(1) << iter->second;

}
cout << endl;

system("pause");
return 0;
}