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
where K is the number of nonzero terms in the polynomial, Ni and aNi (,) 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
题解:用C++写了两次只通过了部分数据额,最后看别人的博客是用C语言来写的,确实,用C++来控制输入输出的格式不太好控制。
#include<iostream> using namespace std; int main() {
float a[1001] = {0};
int k;
int exp;
float coe; cin >> k;
for(int i = 0; i < k; ++i) {
cin >> exp >> coe;
a[exp] += coe;
} cin >> k;
for(int i = 0; i < k; ++i) {
cin >> exp >> coe;
a[exp] += coe;
} int count = 0;
for (int i = 0; i < 1001; ++i) {
if (a[i] != 0) count++;
} cout << count;
for (int i = 1000; i >= 0; --i) {
if (a[i] > 0)
cout << " " << i << " " << a[i];
} }
2021-01-28
Python列表表达式:[ expression for i in iterable ]
poly = [0 for _ in range(1001)]按照习惯,有时候单个独立下划线是用作一个名字,来表示某个变量是临时的或无关紧要的。
例如,在下面的循环中,我们不需要访问正在运行的索引,我们可以使用“_”来表示它只是一个临时值
poly = [0 for _ in range(1001)] def add():
global poly
line = input().split()[1:]
i = 0
while i < len(line) - 1:
poly[int(line[i])] += float(line[i + 1])
i += 2 add()
add() count = 0
for i in range(1000, -1, -1):
if poly[i] != 0:
count += 1 print(count, end='')
for i in range(1000, -1, -1):
if poly[i] != 0:
print(" %d %.1f" % (i, poly[i]), end='')
1002 A+B for Polynomials (25分)的更多相关文章
- 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 ...
- PAT Advanced 1002 A+B for Polynomials (25 分)(隐藏条件,多项式的系数不能为0)
This time, you are supposed to find A+B where A and B are two polynomials. Input Specification: Each ...
- 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 ...
- 【PAT甲级】1002 A+B for Polynomials (25 分)
题意:给出两个多项式,计算两个多项式的和,并以指数从大到小输出多项式的指数个数,指数和系数. AAAAAccepted code: #include<bits/stdc++.h> usin ...
- 1002 A+B for Polynomials (25分) 格式错误
算法笔记上能踩的坑都踩了. #include<iostream> using namespace std; float a[1001];//至少1000个位置 int main(){ in ...
- P1002 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 polyn ...
- PAT 1009 Product of Polynomials (25分) 指数做数组下标,系数做值
题目 This time, you are supposed to find A×B where A and B are two polynomials. Input Specification: E ...
- 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 ...
- 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 ...
随机推荐
- SpringBoot(九):SpringBoot集成Mybatis
(1)新建一个SpringBoot工程,在pom.xml中配置相关jar依赖 贴代码: <!--加载mybatis整合springboot--> <dependency> &l ...
- 细说MySQL连接查询:内连、左连和右连
转: 细说MySQL连接查询:内连.左连和右连 简介: MySQL 的连接查询,通常都是将来自两个或多个表的行结合起来,基于这些表之间的共同字段,进行数据的拼接.首先,要确定一个主表作为结果集,然后将 ...
- (报错解决)Exception encountered during context initialization
转: (报错解决)Exception encountered during context initialization 关键词 JavaEE JavaWeb eclipse XML AspectJ ...
- 从客流统计到营销赋能,Re-ID加速实体商业数字化转型 | 爱分析洞见
2020年中国实体商业受到突发疫情的重大影响.以危机为契机,实体商业加速数字化转型,利用创新应用服务自身业务.在此阶段,基于Re-ID(Person Re-identification,即行人再识别) ...
- rest framework parsers
解析器 机交互的Web服务更倾向于使用结构化的格式比发送数据格式编码的,因为他们发送比简单的形式更复杂的数据 -马尔科姆Tredinnick,Django开发组 REST框架包含许多内置的解析器类,允 ...
- springboot2.0全局异常处理,文件上传过大会导致,方法被执行两次,并且连接被重置
最后发现是内嵌tomcat也有文件大小限制,默认为2MB,我上传的是4MB,然后就炸了.在application.properties中添加server.tomcat.max-swallow-size ...
- 【odoo14】第三章、创建插件
现在我们已经有了开发环境并了解了如何管理实例及数据库,现在让我们来学习下如何创建插件模块. 本章内容如下: 创建和安装模块 完成manifest文件 组织模块文件结构 添加模型 添加菜单及视图 添加访 ...
- Tomcat详解系列(2) - 理解Tomcat架构设计
Tomcat - 理解Tomcat架构设计 前文我们已经介绍了一个简单的Servlet容器是如何设计出来,我们就可以开始正式学习Tomcat了,在学习开始,我们有必要站在高点去看看Tomcat的架构设 ...
- [换根DP]luogu P3647 [APIO2014]连珠线
题面 https://www.luogu.com.cn/problem/P3647 不重复地取树中相邻的两条边,每次得分为两条边权和,问最大得分 分析 容易想到状态 f[i][0/1] 分别表示 i ...
- 开源的 Switch 模拟器——GitHub 热点速览 v.21.12
作者:HelloGitHub-小鱼干 脸滚键盘操作选手小鱼干这里要推荐一个超酷 Switch 模拟器,不能埋没你的游戏天赋.Ryujinx 是一个 C# 写的 Switch 模拟器,1700+ 游戏可 ...