1009. Product of Polynomials (25)

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本文介绍了一种计算两个多项式相乘的算法，并通过C++代码实现了该算法。输入包含两个多项式的系数和指数，输出为这两个多项式相乘后的结果。程序使用了二维循环来完成乘法操作，并确保输出格式与输入格式一致。

1009. Product of Polynomials (25)
时间限制 400 ms 内存限制 65536 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

#include <iostream>
#include <fstream>
using namespace std;

int main()
{
#ifdef _DEBUG
    ifstream cin("data.txt");
#endif // _DEBUG

    const int pro = 1001;
    const int MaxSize = 2 * pro;
    int n, exp, c = 0;
    double Poly1[pro] = { 0 }, poly2[pro] = { 0 };
    double res[MaxSize] = { 0 };
    cin >> n;
    while (n--)
    {
        cin >> exp;
        cin >> Poly1[exp];
    }
    cin >> n;
    while (n--)
    {
        cin >> exp;
        cin >> poly2[exp];
    }
    for (int i = 0; i < pro; ++i)
    {
        for (int j = 0; j < pro; j++)
        {
            res[i+j] += Poly1[i] * poly2[j];
        }
    }
    for (int i = MaxSize - 1; i >= 0; --i)
    {
        if (res[i] !=0.0)
            ++c;
    }
    printf("%d", c);
    if (c == 0)
        printf(" 0 0");
    for (int i = MaxSize - 1; i >= 0; --i)
    {
        if (res[i]!=0.0)
            printf(" %d %.1f", i, res[i]);
    }
#ifdef _DEBUG
    cin.close();
#ifndef _CODEBLOCKS
    system("pause");
#endif // !_CODEBLOCKS
#endif // _DEBUG

    return 0;
}