1009. Product of Polynomials (25)

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 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

和1002题非常类似，仍然是利用hash思想存储AB两个多项式，找出其中的非零项相乘并合并到结果数组result中的。值得注意的是result的最大指数是2000，而不是加法中的1000。

PS：可以根据读入的最大指数去分配数组的内存，减少占用空间。

import java.util.*;

public class Main {

    public static void main(String[] args) {
        float[] A = new float[1001];
        float[] B = new float[1001];
        float[] result = new float[2001];
        Scanner in = new Scanner(System.in);
        int K1 = in.nextInt();
        for (int i = 0; i < K1; i++) {
            A[in.nextInt()] = in.nextFloat();
        }
        int K2 = in.nextInt();
        for (int i = 0; i < K2; i++) {
            B[in.nextInt()] = in.nextFloat();
        }
        for (int i = 0; i < 1001; i++) {
            if (A[i] != 0) {
                for (int j = 0; j < 1001; j++) {
                    if (B[j] != 0) {
                        result[i + j] += A[i] * B[j];
                    }
                }
            }
        }

        StringBuilder sb = new StringBuilder();
        int count = 0;
        for (int i = 2000; i >= 0; i--) {
            if (result[i] != 0) {
                count++;
                sb.append(String.format(" %d %.1f", i, result[i]));
            }
        }
        System.out.println(count + sb.toString());

    }
}