PAT A 1002. A+B for Polynomials

原题：

This time, you are supposed to find A+B where A and B are two polynomials.

Input

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, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10，0 <= NK < … < N2 < N1 <=1000.

Output

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.

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

import java.util.Scanner;

public class Pat1002{

public static void main(String args[]){
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int[] a1 = new int
;
double[] a2 =  new double
;
for (int i = 0; i < n; i++){
a1[i] = input.nextInt();
a2[i] = input.nextDouble();
}
int m = input.nextInt();
int[] b1 = new int[m];
double[] b2 =  new double[m];
for (int i = 0; i < m; i++){
b1[i] = input.nextInt();
b2[i] = input.nextDouble();
}

int i = 0;
int j = 0;
int count = 0;
int count1 = a1[i];
int count2 = b1[j];
String c = "";
while(count1 != -1 || count2 != -1){
if (count1 > count2){
c = c + " " + count1 + " " + Math.round(a2[i] * 10) / 10.0; //保留一位小数
count++;
if (i < a1.length - 1)
count1 = a1[++i];
else
count1 = -1;
}
else if (count1 < count2){
c = c + " " + count2 + " " + Math.round(b2[j] * 10) / 10.0;
count++;
if (j < b1.length - 1)
count2 = b1[++j];
else
count2 = -1;
}
else{
double sum = Math.round(((a2[i] + b2[j]) * 10)) / 10.0;
if (sum != 0){      //某项被消除的情况
c = c + " " + count1 + " " + sum;
count++;
}
if (i < a1.length - 1)
count1 = a1[++i];
else
count1 = -1;
if (j < b1.length - 1)
count2 = b1[++j];
else
count2 = -1;
}
}
if (c == "") //相加为0的情况
c = "0";
else
c = count + c;
System.out.print(c);
input.close();
}
}