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 N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (,) are the exponents and coefficients, respectively. It is given that 1,0.

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.

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
#include<iostream>
using namespace std;
int main(){
int m,n,i,temp,count=0;
float a[1002],b[1002];
cin>>m;
for(i=0;i<m;i++){
cin>>temp;
cin>>a[temp];
}
cin>>n;
for(i=0;i<n;i++){
cin>>temp;
cin>>b[temp];
if(a[temp]==0){
a[temp]=b[temp];
}
else{
a[temp]+=b[temp];
}
}
for(i=0;i<1001;i++){
if(a[i]!=0)
count++;
}
cout<<count;
for(i=1000;i>=0;i--){
if(a[i]!=0){
printf(" %d %.1f",i,a[i]);
}
}
return 0;
}

  

A+B for Polynomials的更多相关文章

  1. 1002. A+B for Polynomials (25)

    题目链接:https://www.patest.cn/contests/pat-a-practise/1002 原题如下: This time, you are supposed to find A+ ...

  2. PAT (Advanced Level) Practise:1002. A+B for Polynomials

    [题目链接] This time, you are supposed to find A+B where A and B are two polynomials. Input Each input f ...

  3. \(\S1 \) Gaussian Measure and Hermite Polynomials

    Define on \(\mathbb{R}^d\) the normalized Gaussian measure\[ d \gamma(x)=\frac{1}{(2\pi)^{\frac{d}{2 ...

  4. 1002. A+B for Polynomials

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

  5. Legendre polynomials

    In mathematics, Legendre functions are solutions to Legendre's differential equation: In particular, ...

  6. PAT 解题报告 1009. Product of Polynomials (25)

    This time, you are supposed to find A*B where A and B are two polynomials. Input Specification: Each ...

  7. 【PAT】1009. Product of Polynomials (25)

    题目链接:http://pat.zju.edu.cn/contests/pat-a-practise/1009 分析:简单题.相乘时指数相加,系数相乘即可,输出时按指数从高到低的顺序.注意点:多项式相 ...

  8. 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 ...

  9. PAT1009:Product of Polynomials

    1009. Product of 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, Yu ...

  10. PAT1002:A+B for Polynomials

    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 ...

随机推荐

  1. java构建简单的HTTP服务器

    使用Java技术构建Web应用时, 我们通常离不开tomcat和jetty之类的servlet容器,这些Web服务器功能强大,性能强劲,深受欢迎,是运行大型Web应用的必备神器. 虽然Java的设计初 ...

  2. lis框架各种方法的使用

    //这个必须是lpedorapp表的主键才行LPEdorAppDB tLPEdorAppDB = new LPEdorAppDB();tLPEdorAppDB.setEdorAcceptNo(mEdo ...

  3. redis提权

    介绍:Redis是一个开源的使用ANSI C语言编写.遵守BSD协议.支持网络.可基于内存亦可持久化的日志型.Key-Value数据库,并提供多种语言的API.它通常被称为数据结构服务器,因为值(va ...

  4. 虚拟机Linux系统ip查询失败问题

    当用SSH连接Linux需要ip地址,但是不论是通过ipconfig命令,还是通过ip addr命令都无法获取Linux的ip,通过以下方法成功解决了该问题: 1.点击编辑里面的虚拟网络编辑器出现如下 ...

  5. 16-ESP8266 SDK开发基础入门篇--TCP 服务器 非RTOS运行版,串口透传(串口回调函数处理版)

    https://www.cnblogs.com/yangfengwu/p/11105466.html 其实官方给的RTOS的版本就是在原先非RTOS版本上增加的 https://www.cnblogs ...

  6. 结构体&文件

    1.本章学习内容总结 1.1学习内容总结 什么是结构类型? 结构Structure类型是一种允许程序员把一些数据分量聚合成一个整体的数据类型. 结构和数组的区别? 结构和数组的最大区别是数组中所有元素 ...

  7. centos7 安装postgresql11

    1 进入postresql官网下载页面,提示了centos相关下载安装等信息. https://www.postgresql.org/download/linux/redhat/   image.pn ...

  8. es6中class类的静态方法、实例方法、实例属性、(静态属性)

    关于类有两个概念,1,类自身,:2,类的实例对象 总的来说:静态的是指向类自身,而不是指向实例对象,主要是归属不同,这是静态属性的核心. 难点1:静态方法的理解 class Foo { static ...

  9. 设置多个className

    有时候我们需要有选择地设置多个className function myComponent(props) { const myClassName = { 'aaa', {'bbb': props.ne ...

  10. mysql tan() 函数

    mysql> ); +--------------------+ | tan(pi()/) | +--------------------+ | 0.9999999999999999 | +-- ...