A+B for Polynomials
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
#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
- 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+ ...
- 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 ...
- \(\S1 \) Gaussian Measure and Hermite Polynomials
Define on \(\mathbb{R}^d\) the normalized Gaussian measure\[ d \gamma(x)=\frac{1}{(2\pi)^{\frac{d}{2 ...
- 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 ...
- Legendre polynomials
In mathematics, Legendre functions are solutions to Legendre's differential equation: In particular, ...
- 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 ...
- 【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 ...
- 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 ...
- 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 ...
- 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
- python 批量打印PDF
有一批PDF文件,好几百个,每个只打印第2,3页,双面打印. 网上搜索一波,方案如下: 安装Ghostscript,GhostView,使用gsprint命令打印pdf文件. gsprint命令参数说 ...
- windbg在加载模块时下断点
假设我们希望在加载特定的dll时中断调试器,例如,我想启用一些SOS命令,而clr还没有加载,当您遇到程序中过早发生的异常,并且您不能依赖手动尝试在正确的时间中断时,这尤其有用.例如,在将调试器附加到 ...
- Redis的订阅、事务、持久化
1.Redius的订阅: 运用关键字subscribe订阅: 关键字publish发布: 发布后,订阅的页面才会出现发布的内容. 2.Redis事务: Redis事务与mysql的事务不同,mysql ...
- 20-ESP8266 SDK开发基础入门篇--C# TCP客户端编写 , 加入数据通信
https://www.cnblogs.com/yangfengwu/p/11192594.html 自行调整页面 连接上以后主动发个数据 namespace TCPClient { public p ...
- shell脚本编程基础之for循环
循环结构 循环需要有进入条件和退出条件,如果没有退出条件,则就会一直循环下去 for 变量 in 列表:do 循环体 done 生成列表及示例 {1..100}:生成1到100的整数列表 `comma ...
- Myschool试题
题目: 1.查询所有学生记录,包含年级名称2.查询S1年级下的学生记录 一.com.myschool.dao 1 BaseDao package com.myschool.dao; import ja ...
- 「ZJOI2019」Minmax搜索
传送门 Solution 叶子节点的变化区间是连续的,可得知非叶子节点的权值变化区间也是连续的 由此可知,\(W\)的变化值的可行域也是连续的,所以只需要看它能否变为\(W+1\)或\(W-1\) 对 ...
- element ui 中的时间选择器怎么设置默认值/el-date-picker区间选择器怎么这是默认值
template代码 <el-date-picker value-format="yyyy-MM-dd" v-model="search.date" ty ...
- (三)Cisco dhcp snooping实例1-单交换机(DHCP服务器和DHCP客户端位于同一VLAN)
环境:cisco dhcp server和客户端都属于vlan27,dhcp server 接在交换机G0/1,客户端接在交换机的G0/2 cisco dhcp server相关配置 ip dhcp ...
- JdkDynamicAopProxy与CglibAopProxy介绍
继续上一篇的介绍 1.上一篇分析到createAopProxy方法,创建Aop代理对象 protected final synchronized AopProxy createAopProxy() { ...