https://pintia.cn/problem-sets/994805342720868352/problems/994805526272000000

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

代码:

#include <bits/stdc++.h>
using namespace std; double a[1010], b[1010];
int A, B;
double sum[1010]; int main() {
scanf("%d", &A);
for(int i = 1; i <= A; i ++) {
int x;
scanf("%d", &x);
scanf("%lf", &a[x]);
}
scanf("%d", &B);
for(int i = 1; i <= B; i ++) {
int y;
scanf("%d", &y);
scanf("%lf", &b[y]);
} int cnt = 0;
for(int i = 0; i <= 1000; i ++) {
if(a[i] && b[i])
sum[i] = a[i] + b[i];
else if(a[i])
sum[i] = a[i];
else if(b[i])
sum[i] = b[i];
}
for(int i = 0; i <= 1000; i ++) {
if(sum[i])
cnt ++;
}
printf("%d", cnt);
for(int i = 1000; i >= 0; i --) {
if(sum[i])
printf(" %d %.1lf", i, sum[i]);
}
printf("\n");
return 0;
}

  

PAT 甲级 1002 A+B for Polynomials的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. pat甲级1002

    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】1002. A+B for Polynomials (25)

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

  7. 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 ...

  8. 甲级1002 A+B for Polynomials (25)

    题目描述: This time, you are supposed to find A+B where A and B are two polynomials. Input Each input fi ...

  9. 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 ...

随机推荐

  1. ABAP术语-Error Message

    Error Message 原文:http://www.cnblogs.com/qiangsheng/archive/2008/01/30/1058283.html Information from ...

  2. 详解Linux运维工程师

    运维工程师是从一个呆逼进化为苦逼再成长为牛逼的过程,前提在于你要能忍能干能拼,还要具有敏锐的嗅觉感知前方潮流变化.如:今年大数据,人工智能比较火……(相对表示就是 Python 比较火) 之前写过运维 ...

  3. linux服务基础之ftp服务

    ftp是一种文件传输协议,我们以redhat6.9为服务器系统,来介绍一下ftp服务器,这里我们先介绍一下ftp协议工作的原理 ftp协议可以在不同类型的计算机之间传输文件,工作流程大致为 1:客户机 ...

  4. jQuery(一)初识

    jQuery 的功能概括 1.html 的元素选取 2.html的元素操作 3.html dom遍历和修改 4.js特效和动画效果 5.css操作 6.html事件操作 7.ajax异步请求方式 se ...

  5. U盘被分区后恢复方法

    一:运行cmd 二:输入diskpart,按enter. 三:输入list disk,按enter. 四:选择优U盘,输入select disk X(X代表磁盘后面的数字0.1,可磁盘的大小来判断数字 ...

  6. 6-C++远征之封装篇[上]-学习笔记

    C++远征之封装篇(上) 课程简介 类(抽象概念),对象(真实具体) 配角: 数据成员和成员函数(构成了精彩而完整的类) 构造函数 & 析构函数(描述了对象的生生死死) 对象复制和对象赋值 ( ...

  7. 贪心算法之Prim

    Prim与Dijistra算法有异曲同工之妙,只不过Dijistra是求最短路径,每次添加到集合中的是到固定起始点的最短距离,而Prim是求最小生成树,是整个图所有权重的最小和,每次添加到集合中的是到 ...

  8. WCF入门四[WCF的通信模式]

    一.概述 WCF的通信模式有三种:请求/响应模式.单向模式和双工通信. 二.请求/响应模式 请求/响应模式就是WCF的默认模式,前面几篇随笔中的示例都是这种模式,当客户端发送请求后(非异步状态下),即 ...

  9. 8 TFTP代码详解 协议写在程序中

    1.版本1:发送请求 # -*- coding:utf-8 -*- import struct from socket import * #0. 获取要下载的文件名字: downloadFileNam ...

  10. 3122 奶牛代理商 VIII(状压dp)

    3122 奶牛代理商 VIII  时间限制: 3 s  空间限制: 256000 KB  题目等级 : 大师 Master     题目描述 Description 小徐是USACO中国区的奶牛代理商 ...