[PAT甲级]1009. Product of Polynomials (25)(求多项式的积)

1009. Product of Polynomials (25)

原题链接
相似题目 1002. A+B for Polynomials (25)(求两个多项式的和)

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

题目大意：

• 给定两个多项式，输出它们的积
• 样例中多项式A有2个项： 2.4 * X^1 + 3.2 * X^0
• 样例中多项式B有2个项： 1.5 * X^2 + 0.5 * X^1
• A*B有3个项 = 3.6 * X^3 + 6.0 * X^2+ 1.6 * X^1
• X^2 表示 X的平方,最好是自己手写计算一下，指数相加，系数相乘
• 精确到小数点后一位

代码：

#include <iostream>
#include <vector>
#include <cstdio>
using namespace std;

int main()
{
    vector<double> v(2001);//存储最终系数，指数最大为1000，A*B最大指数是2000
    int m;
    cin >> m;
    vector<int> NK;//指数
    vector<double> ank;//系数
    for(int i=0; i<m; i++){
        int a;
        double b;
        cin >> a >> b;
        NK.push_back(a);
        ank.push_back(b);
    }
    int n;
    cin >> n;
    for(int i=0; i<n; i++){
        int a;//指数
        double b;//系数
        cin >> a >> b;
        for(int j=0; j<ank.size(); j++){
            v[a+NK[j]] += b*ank[j];
        }
    }
    vector<int> res;//保存要输出的指数
    for(int i=v.size()-1; i>=0; i--){
        if(v[i] != 0)
            res.push_back(i);
    }
    cout << res.size();
    for(int i=0; i<res.size(); i++)
        printf(" %d %.1f", res[i], v[res[i]]);
    return 0;
}