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  • 【JAVA】1002 A+B for Polynomials (25分) PAT甲级 PAT (Advanced Level) Practice

    1002 A+B for 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 N1N_1 aN1a_N1 N2N_2 aN2a_N2NKN_K aNKa_NK
    每个输入文件包含一个测试用例。 每个案例占用2行,并且每行包含一个多项式的信息:
    K N1N_1 aN1a_N1 N2N_2 aN2a_N2NKN_K aNKa_NK

    where K is the number of nonzero terms in the polynomial, NiN_iandaNia_Ni(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10,0≤NKN_K​<⋯<N2N_2<N1N_1≤1000.
    其中K是多项式中非零项的数量,NiN_iandaNia_Ni(i=1,2,⋯,K) 分别是指数和系数。 假设1≤K≤10,0≤NKN_K​<⋯<N2N_2<N1N_1≤1000.

    Output Specification:
    For each test case you should output the sum 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 to 1 decimal place.
    对于每个测试用例,应在一行中输出A和B的总和,格式与输入相同。 请注意,每行末尾不得有多余的空间。 请精确到小数点后一位。

    Sample Input:
    2 1 2.4 0 3.2
    2 2 1.5 1 0.5

    Sample Output:
    3 2 1.5 1 2.9 0 3.2

    要点

    1. 最难理解的是题目
      在这里插入图片描述

    2. 请注意,每行末尾不得有多余的空间。

    3. 请精确到小数点后一位。

      System.out.printf("%.1f", C[i]);
      
    4. 注意一个特殊样例,比如输入两行分别是
      1 2 1
      1 2 -1
      这就是x2x^2+x2-x^2,输出为0

    代码

    import java.util.Arrays;
    import java.util.Scanner;
    
    public class PolynomialAddition1 {
        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            Double[] A = new Double[1001];
            for (int i = 0; i < 1001; i++) { //防止 C[i] = A[i] + B[i];发生空指针异常
                A[i] = 0.0;
            }
            Double[] B = new Double[1001];
            for (int i = 0; i < 1001; i++) {//防止 C[i] = A[i] + B[i];发生空指针异常
                B[i] = 0.0;
            }
    
            int K = sc.nextInt();
            int index;
            while (K != 0) {
                index = sc.nextInt(); //指数
                A[index] = sc.nextDouble(); //系数
                K--;
            }
    //        System.out.println(Arrays.toString(A));
    
            K = sc.nextInt();
            while (K != 0) {
                index = sc.nextInt(); //指数
                B[index] = sc.nextDouble(); //系数
                K--;
            }
            sc.close();
    //        System.out.println(Arrays.toString(B));
    
            double[] C = new double[1001];
            int count = 0;
            /*//错误
            for (int i = 0; i < 1001; i++) {
                if (A[i] != 0 || B[i] != 0) {
                //这得小心一种情况。输入的两行分别是 a:1 1 1  b:1 1 -1。
                //A[1]+B[1]=0,这时count不应该再加
                    count++;
                    C[i] = A[i] + B[i];
                }
            }*/
            //改正
            for (int i = 0; i < 1001; i++) {
                C[i] = A[i] + B[i];
                if ((A[i] != 0 || B[i] != 0) && C[i] != 0) {
                    // 这得小心一种情况。输入的两行分别是 a:1 1 1  b:1 1 -1。
                    // A[1]+B[1]=0,这时count不应该再加
                    count++;
                }
            }
    //        System.out.println(Arrays.toString(C));
            System.out.print(count);
            for (int i = 1000; i >= 0; i--) {
                if (C[i] != 0) {
                    System.out.print(" " + i + " ");
                    System.out.printf("%.1f", C[i]);
                }
            }
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/DreamingFishZIHao/p/12982957.html
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