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.LinkedList;
import java.util.Queue;
import java.util.Scanner; public class Main {
public static void main(String[] args) {
Queue<A> q1 = new LinkedList<A>();
Queue<A> q2 = new LinkedList<A>();
Queue<A> q3 = new LinkedList<A>();
Scanner input = new Scanner(System.in);
int n = input.nextInt();
for(int i=0;i<n;i++){
A a = new A();
a.a = input.nextInt();
a.b = input.nextDouble();
q1.add(a);
}
int m = input.nextInt();
for(int i=0;i<m;i++){
A a = new A();
a.a = input.nextInt();
a.b = input.nextDouble();
q2.add(a);
}
while(!q1.isEmpty()&&!q2.isEmpty()){
int a1 = q1.peek().a;
double b1 = q1.peek().b;
int a2 = q2.peek().a;
double b2 = q2.peek().b;
A a = new A();
if(a1==a2){
a.a = a1;
a.b = b1+b2;
q1.poll();
q2.poll();
}if(a1>a2){
a.a = a1;
a.b = b1;
q1.poll();
}if(a1<a2){
a.a = a2;
a.b = b2;
q2.poll();
}
if(a.b!=0)
q3.add(a);
}
while(!q1.isEmpty()){
q3.add(q1.poll());
}
while(!q2.isEmpty()){
q3.add(q2.poll());
}
System.out.print(q3.size()); while(!q3.isEmpty()){
System.out.print(" "+q3.peek().a+" ");
System.out.printf("%.1f",q3.peek().b);
q3.poll();
} }
}
class A{
int a;
double b;
}

PAT 1002. A+B for Polynomials (25)的更多相关文章

  1. PAT 1002. A+B for Polynomials (25) 简单模拟

    1002. A+B for Polynomials (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue T ...

  2. PAT 1002 A+B for Polynomials (25分)

    题目 This time, you are supposed to find A+B where A and B are two polynomials. Input Specification: E ...

  3. PAT (Advanced Level) Practice 1002 A+B for Polynomials (25 分) 凌宸1642

    PAT (Advanced Level) Practice 1002 A+B for Polynomials (25 分) 凌宸1642 题目描述: This time, you are suppos ...

  4. 【PAT】1002. A+B for Polynomials (25)

    1002. A+B for Polynomials (25) This time, you are supposed to find A+B where A and B are two polynom ...

  5. PAT甲级 1002 A+B for Polynomials (25)(25 分)

    1002 A+B for Polynomials (25)(25 分) This time, you are supposed to find A+B where A and B are two po ...

  6. PAT 甲级1002 A+B for Polynomials (25)

    1002. A+B for Polynomials (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue T ...

  7. PAT甲 1002. A+B for Polynomials (25) 2016-09-09 22:50 64人阅读 评论(0) 收藏

    1002. A+B for Polynomials (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue T ...

  8. 1002 A+B for Polynomials (25)(25 point(s))

    problem 1002 A+B for Polynomials (25)(25 point(s)) This time, you are supposed to find A+B where A a ...

  9. pat 1002 A+B for Polynomials (25 分)

    1002 A+B for Polynomials (25 分) This time, you are supposed to find A+B where A and B are two polyno ...

随机推荐

  1. CentOS 7中mariadb编译安装教程systemctl启动

    mariadb做为mysql的替代品 现在centos的新版本yum包已换成mariadb 跟上篇一样只是启动方式改为systemd 安装一些库 yum install gcc gcc-c++ wge ...

  2. ccentos 7下安装php5.6并使用nginx + php-fpm部署多个不同端口网站

    作为一个的勤杂工,近期因公司内部信息化的需求,给新进员工提供基础的知识培训和介绍,也为了给公司内部建立一个沟通交流的平台,百度找了开源的百科系统HDwiki和开源的问答系统Tipask问答系统,蛋痛的 ...

  3. 转-Cannot refer to an instance field arg while explicitly invoking a constructor

    编译失败: Cannot refer to an instance field arg while explicitly invoking a constructor  调用方法时不能引用一个实例变量 ...

  4. 如何查询当前手机的cpu架构,so库导入工程又出异常了?

    执行adb命令: adb shell cat /proc/cpuinfo 对应文件夹 AArch64 == arm64-v8a ARMv7 == armeabi-v7a ............等 其 ...

  5. [浪风转载]Jquery取得iframe中元素的几种方法

    iframe在复合文档中经常用到,利用jquery操作iframe可以大幅提高效率,这里收集一些基本操作 DOM方法:父窗口操作IFRAME:window.frames["iframeSon ...

  6. sqoop1.99.3配置过程

    感谢:http://www.myexception.cn/database/1600202.html 官网配置指导:http://sqoop.apache.org/docs/1.99.3/Instal ...

  7. il code swtich

    public enum RangeItemType { CompanyPicker = 1, //公司STR_COMPANY BrDivPicker = 2, //分行STR_BRDIV Depart ...

  8. JAVA的protected权限

    1.派生类能够訪问父类的protected数据.这是毫无疑问的. 2.今天做Think in java的习题发现,同一个包内的一个类调用还有一个类的protected数据是能够的.代码例如以下: pa ...

  9. Android Otto框架浅析

    今天要介绍的是一个Android中使用得比較多的android 事件总线 EventBus模式的一个框架Otto. Otto 官网:http://square.github.io/otto/ 一.An ...

  10. Android-ViewPagerIndicator框架使用——LinePageIndicator

    前言:LinePageIndicator类似CirclePageIndicator,只是将圆点指示变成了长条指示. 一:使用是定义的布局文件simple_lines : <LinearLayou ...