1009. Product of polynomials (25) -- Pat (Advanced Level) practise

Question information:
1009. Product of polynomials (25) Time Limit 400 MS
The memory limit is 32000 kb.
Code length limit: 16000 B
Criterion author Chen, Yue

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_{N1 N2N2... NKNK, where k is the number of nonzero terms in the polynomial, Ni andNi (I = 1, 2 ,..., k) are the exponents and coefficients, respectively. it is given that 1 <= k <= 10, 0 <= NK <... <N1 <= 1000.}

Output specification:

For each test case you shoshould 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 3.22 2 1.5 1 0.5

Sample output

3 3 3.6 2 6.0 1 1.6

The Code is as follows:

#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