PAT(甲级)1009 Product of Polynomials （25 分)

最新推荐文章于 2022-02-20 15:19:12 发布

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最新推荐文章于 2022-02-20 15:19:12 发布

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本文详细介绍了一种解决多项式相乘问题的算法，通过结构体存储多项式的系数和指数，实现了两个多项式的精确相乘，并提供了完整的C++代码实现。

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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

简析：这道题是多项式的相乘，因为同时设计系数和指数，所以设置一个结构体存放系数和指数，然后每次相乘都进行变化并放入答案数组。（貌似是暴力解法）

#include<iostream>
using namespace std;
struct Poly{
    int exp;
    double cof;
} poly[1001];
double ans[2001] = {0};
int main()
{
    int n, m, num = 0;
    cin>>n;
    for (int i = 0; i < n;i++)
    {
        cin >> poly[i].exp >> poly[i].cof;
    }
    cin >> m;
    for(int i = 0; i < m;i++)
    {
        int exp;
        double cof;
        cin >> exp >> cof;
        for (int j = 0; j < n; j++)
        {
            ans[exp + poly[j].exp] += cof * poly[j].cof;
        }
    }
    for (int i = 0; i < 2001;i++)
    {
        if(ans[i] != 0.0)
            num++;
    }
    cout << num;
    for (int i = 2000; i >= 0;i--)
    {
        if(ans[i] != 0.0)
            printf(" %d %.1f",i,ans[i]);
    }
    return 0;
}