【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_1N1 aN1a_N1aN1 N2N_2N2 aN2a_N2aN2… NKN_KNK aNKa_NKaNK
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_1N1 aN1a_N1aN1 N2N_2N2 aN2a_N2aN2… NKN_KNK aNKa_NKaNK
where K is the number of nonzero terms in the polynomial, NiN_iNiandaNia_NiaNi(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10,0≤NKN_KNK<⋯<N2N_2N2<N1N_1N1≤1000.
其中K是多项式中非零项的数量,NiN_iNiandaNia_NiaNi(i=1,2,⋯,K) 分别是指数和系数。 假设1≤K≤10,0≤NKN_KNK<⋯<N2N_2N2<N1N_1N1≤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
-
最难理解的是题目
-
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。
System.out.printf("%.1f", C[i]);
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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^2x2+−x2-x^2−x2,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]);} } } }爱做梦的鱼 原创文章 169获赞 1083访问量 7万+ 关注 私信
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