1009_Product_of_Polynamials

1009. Product of Polynomials (25)

时间限制
400 ms
内存限制
32000 kB
代码长度限制
16000 B
判题程序
Standard
作者
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 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

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解：理解怎么乘后其实跟1002类似，只是存结果的数组需开得更大

#include <iostream>
#include <iomanip>
using namespace std;
 
int main()
{
    int m,n;
    double m_mul[1001];
    double n_mul[1001];
    double product[2001];
 
    for(int i = 0; i < 1001; ++i){
        m_mul[i] = n_mul[i] = product[i] = 0.f;
    }
    cin >> m;
    for(int i = 0; i < m; ++i){
        int index1;
        cin >> index1;
        cin >> m_mul[index1] ;
    }
    cin >> n;
    for(int i = 0; i < n; ++i){
        int index2;
        cin >> index2;
        cin >> n_mul[index2];
    }
 
    for(int i = 0; i < 1001; ++i)
        for(int j = 0; j < 1001; ++j){
            if(m_mul[i] != 0 && n_mul[j] != 0){
                product[i + j] +=m_mul[i] * n_mul[j];
            }
        }
 
    int num = 0;
    for(int i = 0; i < 2001; ++i)
        if(product[i] != 0)
            num++;
 
    cout << num;
    for(int i = 2000; i >= 0; --i){
        if(product[i] != 0.f)
            cout << ' ' << i << ' ' << setiosflags(ios::fixed) << setprecision(1) <<  product[i];
    }
    return 0;
}