标签:25 PAT Level int index System sc 1001 out
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 N1 aN1 N2 aN2… NK aNK
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 N1 aN1 N2 aN2… NK aNK
where K is the number of nonzero terms in the polynomial, NiandaNi(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10,0≤NK<⋯<N2<N1≤1000.
其中K是多项式中非零项的数量,NiandaNi(i=1,2,⋯,K) 分别是指数和系数。 假设1≤K≤10,0≤NK<⋯<N2<N1≤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]);
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;
for (int i = 0; i < 1001; i++) {
if (A[i] != 0 || B[i] != 0) {
count++;
C[i] = A[i] + B[i];
}
}
// 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]);
}
}
}
}
标签:25,PAT,Level,int,index,System,sc,1001,out 来源: https://blog.csdn.net/weixin_43124279/article/details/105823853
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