题目

1009 Product of Polynomials (25分)

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 a**N1 N2 a**N2 ... N**K aNK

where K is the number of nonzero terms in the polynomial, N**i and aNi (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N**K<⋯<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

理解与算法

多项式乘法，思路也不难，遍历两个多项式，相乘即可，M*N的复杂度，需要考虑的是存储方式。

题目里给出的范围是1-1000，所以数组大小设置为2001就足够了，但是，如果指数范围非常大，大到几万，几十万，就不太好用数组来实现，可以考虑用map来实现。

代码与实现

//
// Created by tanknee on 2021/1/24.
//

#include <iostream>
#include <map>

using namespace std;

int main() {
    int count, exponents;
    double coefficients, m1[1001] = {0.0}, result[2001] = {0.0};
    cin >> count;
    for (int i = 0; i < count; ++i) {
        cin >> exponents >> coefficients;
        m1[exponents] = coefficients;
    }
    cin >> count;
    for (int i = 0; i < count; ++i) {
        cin >> exponents >> coefficients;
        for (int j = 0; j < 1001; ++j) {
            if (m1[j] != 0.0)
                result[j + exponents] += m1[j] * coefficients;
        }
    }
    count = 0;
    for (int i = 2000; i >= 0; i--) {
        if (result[i] != 0.0)
            count++;
    }
    printf("%d", count);
    for (int i = 2000; i >= 0; i--) {
        if (result[i] != 0.0)
            printf(" %d %.1f", i, result[i]);
    }
    return 0;
}

标签：count,Product,PAT,int,0.0,result,1009,exponents,coefficients
来源： https://www.cnblogs.com/tanknee/p/14320813.html