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 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 (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10,0≤NK<⋯<N2<N1≤1000.
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 <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define wzf ((1 + sqrt(5.0)) / 2.0)
#define INF 0x3f3f3f3f
#define eps 0.0000001
#define LL long long
using namespace std; const int MAXN = 1e3 + ;
int cnt = , book[MAXN] = {}, n, a;
double A[MAXN] = {0.0}, B[MAXN] = {0.0}, C[MAXN] = {0.0}, b; int main()
{
freopen("Date1.txt", "r", stdin);
scanf("%d", &n);
while (n --)
{
scanf("%d%lf", &a, &b);
A[a] = b;
}
scanf("%d", &n);
while (n --)
{
scanf("%d%lf", &a, &b);
B[a] += b;
} for (int i = ; i >= ; -- i)
{
if (A[i] != || B[i] != )
C[i] = A[i] + B[i];
if (C[i] != )
++ cnt;
}
printf("%d", cnt);
for (int i = ; i >= ; -- i)
{
if (C[i] != )
printf(" %d %.1f", i, C[i]);
}
printf("\n");
return ;
}
pat 1002 A+B for Polynomials (25 分)的更多相关文章
- 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 (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 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 ...
- 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 ...
- PAT 1002. A+B for Polynomials (25)
This time, you are supposed to find A+B where A and B are two polynomials. Input Each input file con ...
- PAT Advanced 1009 Product of Polynomials (25 分)(vector删除元素用的是erase)
This time, you are supposed to find A×B where A and B are two polynomials. Input Specification: Each ...
- 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 ...
- 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 ...
- 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 ...
随机推荐
- 使用Spring 或Spring Boot实现读写分离( MySQL实现主从复制)
http://blog.csdn.net/jack85986370/article/details/51559232 http://blog.csdn.net/neosmith/article/det ...
- 剑指Offer(十九)——顺时针打印矩阵
题目描述 输入一个矩阵,按照从外向里以顺时针的顺序依次打印出每一个数字. 例如,如果输入如下4 X 4矩阵: 1 2 3 4 5 6 7 8 9 10 11 ...
- jmeter-定时器使用
在一般性能测试过程中,往往在前一个请求之后等待一段时间再执行下一个请求,这时会用到定时器. 以下列举常用的3中: 1.固定定时器: 2.
- C#初始类和命名空间
本节内容: 1.剖析Hello,World程序 1.1初始类(class)与名称空间(namespace) 2.类库的引用 2.1DLL的引用(黑盒引用) 2.2项目引用(白盒引用) 2.3建立自己的 ...
- CodeForces - 1214D D. Treasure Island
题目链接:https://vjudge.net/problem/2728294/origin 思路:可以抽象成管道,先试试能不能找到一个通道能通到终点, 如果可以则封锁这个通道,一个石头即可, 再试试 ...
- maven 打包 spring boot 生成docker 镜像
1.所使用材料 ,spring boot 项目 基于maven ,maven 工具, docker工具 ps:为啥使用 docker 公司微服务需要启动太多,有两个优点吧! 1.方便管理,2.减少服务 ...
- java 连续数字数组分组
问题: 1. 将Lis list = Arrays.asList(1,2,3,5,8,9,10), 拆分成 [1,2,3] .[5]. [8,9,10] , 2. 再传入一个数字 9, 将匹配数字9的 ...
- Mutex vs Semaphore vs Monitor vs SemaphoreSlim
C#开发者(面试者)都会遇到Mutex,Semaphore,Monitor,SemaphoreSlim这四个与锁相关的C#类型,本文期望以最简洁明了的方式阐述四种对象的区别. 线程安全 教条式理解 如 ...
- windows下Eclipse远程连接linux hadoop远程调试 经验(一)
环境 Windows 7 64bit JDK 1.6.0_45 (i586) JDK 1.7.0_51 (i586) Eclipse Kepler Eclipse -plugin-1.2.1.ja ...
- django-ckedit
(转载) 在django项目中使用django-ckeditor 安装django-ckeditor pip install django-ckeditor 安装Pillow Pillow是pyt ...