标签：product of polynomia advanced level practice pat 1009
题目信息：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 Input2 1 2.4 0 3.2 2 2 1.5 1 0.5Sample Output
3 3 3.6 2 6.0 1 1.6
代码如下：
#include <iostream>
#include <map>
using namespace std;
int main()
{
map<int, float> m1, m2, m3;
int k, n;
float a;
cin >> k;
for (int i = 0; i < k; i++)
{
cin >> n >> a;
m1[n] = a;
}
cin >> k;
for (int i = 0; i < k; i++)
{
cin >> n >> a;
m2[n] = a;
}
for (map<int, float>::iterator it1 = m1.begin(); it1 != m1.end(); it1++)
{
for (map<int, float>::iterator it2 = m2.begin(); it2 != m2.end(); it2++)
{
m3[it1->first + it2->first] += it1->second * it2->second;
}
}
cout.setf(ios_base::fixed, ios_base::floatfield);
cout.precision(1);
for (map<int, float>::iterator it = m3.begin(); it != m3.end(); it++)
{
if (it->second == 0.0f)
m3.erase(it);
}
cout << m3.size();
for (map<int, float>::reverse_iterator it = m3.rbegin(); it != m3.rend(); it++)
{
cout <<" "<< it->first << " " << it->second;
}
cout << endl;
return 0;
}
1009. Product of Polynomials (25)——PAT (Advanced Level) Practise
标签：product of polynomia advanced level practice pat 1009
原文地址：http://blog.csdn.net/xianyun2009/article/details/39519067
