PAT Advanced 1009 Product of Polynomials（25）

题目描述：

This time, you are supposed to find A×B where A and B are two polynomials.

Input Specification:

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

算法描述：多项式乘法

题目大意：

给出两个多项式的 项数 和 各个系数，求两多项式相乘之后的多项式

#include<iostream>
using namespace std;
double nk,a[1010] = {0.0},c[2020]={0.0}; // nk 系数   数组下标 次数
int main()
{
    int k1,k2,sum=0,n;
    cin >> k1;
    for(int i = 0 ; i < k1 ; i++)
    {
        cin >> n >> nk ;
            a[n] = nk ;
    }
    cin >> k2;
    for(int i = 0 ; i < k2 ; i++)
    {
        cin >> n >> nk ;
        for(int j = 0 ; j <= 1000 ; j++ )
                c[ n + j ] += nk * a[j] ;
    }
    for(int i = 0 ; i <= 2000 ; i++ )
        if(c[i])
            sum++;
    cout << sum ;
    for(int i = 2000 ; i >= 0 ; i-- )
        if(c[i])
            printf(" %d %.1lf", i, c[i]);
    return 0;
}

标签：25,Product,nk,PAT,int,多项式,cin,k2,k1
来源： https://www.cnblogs.com/yztozju/p/16580454.html