• [PAT]1009 Product of Polynomials (25 分)Java


    This time, you are supposed to find A×BA imes BA×B where AAA and BBB 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:

    KKK N1N_1N1​​aN1a_{N_1}aN1​​​​N2N_2N2​​aN2a_{N_2}aN2​​​​ ... NKN_KNK​​aNKa_{N_K}aNK​​​​

    where KKK is the number of nonzero terms in the polynomial, NiN_iNi​​ and aNia_{N_i}aNi​​​​ (i=1,2,⋯,Ki=1, 2, cdots , Ki=1,2,,K) are the exponents and coefficients, respectively. It is given that 1≤K≤101le K le 101K10, 0≤NK<⋯<N2<N1≤10000 le N_K < cdots < N_2 < N_1 le 10000NK​​<<N2​​<N1​​1000.

    Output Specification:

    For each test case you should output the product of AAA and BBB 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



     1 package pattest;
     2 
     3 import java.util.Scanner;
     4 
     5 /**
     6  * @Auther: Xingzheng Wang
     7  * @Date: 2019/2/18 21:38
     8  * @Description: pattest
     9  * @Version: 1.0
    10  */
    11 public class PAT1009 {
    12     public static void main(String[] args) {
    13         int k1, k2, count = 0;
    14         double[] a = new double[1001];
    15         double[] b = new double[2001];
    16         Scanner sc = new Scanner(System.in);
    17         k1 = sc.nextInt();
    18         for (int i = 0; i < k1; i++) {
    19             int index = sc.nextInt();
    20             a[index] = sc.nextDouble();
    21         }
    22         k2 = sc.nextInt();
    23         for (int i = 0; i < k2; i++) {
    24             int index = sc.nextInt();
    25             double s = sc.nextDouble();
    26             for (int j = 0; j < 1001; j++) {
    27                 if (a[j] != 0) {
    28                     b[index + j] += s * a[j];
    29                 }
    30             }
    31         }
    32         for (int i = 0; i < 2001; i++) {
    33             if (b[i] != 0)
    34                 count++;
    35         }
    36         System.out.print(count);
    37         for (int i = 2000; i >= 0; i--) {
    38             if (b[i] != 0)
    39                 System.out.printf(" %d %.1f", i, b[i]);
    40         }
    41     }
    42 }
  • 相关阅读:
    CNN(卷积神经网络)入门
    基于linux vim环境python代码自动补全
    Linux 基本bash命令
    基于pytorch的CNN、LSTM神经网络模型调参小结
    深度学习中Batch size对训练效果的影响
    argparse.ArgumentParser()用法解析
    大数据学习之Hive数据仓库 20
    centOS中安装MySQL超级方便简单的方法
    大数据学习之zookeeper案例节点动态上下线感知19
    大数据学习之zookeeper客户端的命令行及API操作18
  • 原文地址:https://www.cnblogs.com/PureJava/p/10497968.html
Copyright © 2020-2023  润新知