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

使用map进行映射即可,但需注意输出格式
 #include <iostream>
//#include <vector>
#include <map>
#include <stack>
using namespace std; int main()
{
map<double, double,greater<double>>data;//从大到小排序
for (int i = ; i < ; ++i)//输入两组数据
{
int k;
cin >> k;
for (int j = ; j < k; ++j)//接受每组数据
{
double a, b;
cin >> a >> b;
data[a] += b;
if (data[a] == )//系数为0则删除
data.erase(a);
}
}
cout << data.size();
for (auto ptr = data.begin(); ptr != data.end(); ++ptr)
{
cout << " " << ptr->first << " ";
printf("%.1f", ptr->second);
}
return ;
}
 

PAT甲级——A1002 A+B for Polynomials的更多相关文章

  1. PAT——甲级1009:Product of Polynomials;乙级1041:考试座位号;乙级1004:成绩排名

    题目 1009 Product of Polynomials (25 point(s)) This time, you are supposed to find A×B where A and B a ...

  2. PAT 甲级 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 ...

  3. 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 ...

  4. PAT 甲级 1002 A+B for Polynomials

    https://pintia.cn/problem-sets/994805342720868352/problems/994805526272000000 This time, you are sup ...

  5. 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 ...

  6. PAT甲级——1002 A+B for Polynomials

    PATA1002 A+B for Polynomials This time, you are supposed to find A+B where A and B are two polynomia ...

  7. 【PAT】A1002 A+B for Polynomials

    仅有两个要注意的点: 如果系数为0,则不输出,所以输入结束以后要先遍历确定系数不为零的项的个数 题目最后一句,精确到小数点后一位,如果这里忽略了,会导致样例1,3,4,5都不能通过

  8. 【PAT甲级】1009 Product of Polynomials (25 分)

    题意: 给出两个多项式,计算两个多项式的积,并以指数从大到小输出多项式的指数个数,指数和系数. trick: 这道题数据未知,导致测试的时候发现不了问题所在. 用set统计非零项时,通过set.siz ...

  9. PAT甲级题解(慢慢刷中)

    博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给 ...

随机推荐

  1. Django 分页器模板

    返回链接: djang ORM 分页器模板: class Pagination(object): def __init__(self,current_page,all_count,per_page_n ...

  2. 2019 Multi-University Training Contest 3 B 支配树

    题目传送门 题意:给出衣服有向无环图(DAG),,定义出度为0的点为中心城市,每次询问给出两个点,求破坏任意一个城市,使得这两个点至少有一个点无法到达中心城市,求方案数. 思路:首先建立反向图,将城市 ...

  3. postman在有登录认证的情况下进行接口测试!!!

    1.启动自己的项目之后直接使用浏览器进行登录,登陆之后随意点击一个请求,F12找到该请求中请求头的Cookie键值对. 2.将该键值对复制粘贴到postman中的请求Headers中,如下图. 3,请 ...

  4. 【JZOJ6346】ZYB和售货机

    description analysis 其实这个连出来的东西叫基环内向树 先考虑很多森林的情况,也就是树根连回自己 明显树根物品是可以被取完的,那么买树根的价钱要是儿子中价钱最小的那个 或者把那个叫 ...

  5. html5 js 监听网络在线与离线

    <!doctype html> <html> <head> <meta http-equiv="content-type" content ...

  6. (转)Wireshark "The NPF driver isn’t running…"(

    转:http://blog.sina.com.cn/s/blog_4bfd07180100e3ar.html 前几天重装系统,装上了windows7 RC系统.昨天开始尝试装上了wireshark 这 ...

  7. .net core, docker 在vs2019开发过程中的问题以及解决办法

    .net core, docker 在vs2019开发过程中的问题以及解决办法 记录下来,帮助Ta人~ 1.vs调试,快Build完后提示Docker 端口:xxxx,xxxx,xxxx占用 解决办法 ...

  8. PAT甲级——A1131 Subway Map【30】

    In the big cities, the subway systems always look so complex to the visitors. To give you some sense ...

  9. URL类发送请求

    import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;import jav ...

  10. Spring - 框架入门

    认识 Spring 框架 Spring 框架是 Java 应用最广的框架,它的成功来源于理念,而不是技术本身,它的理念包括 IoC (Inversion of Control,控制反转) 和 AOP( ...