1058 A+B in Hogwarts (20分)(水)
If you are a fan of Harry Potter, you would know the world of magic has its own currency system -- as Hagrid explained it to Harry, "Seventeen silver Sickles to a Galleon and twenty-nine Knuts to a Sickle, it's easy enough." Your job is to write a program to compute A+B where A and B are given in the standard form of Galleon.Sickle.Knut (Galleon is an integer in [0], Sickle is an integer in [0, 17), and Knut is an integer in [0, 29)).
Input Specification:
Each input file contains one test case which occupies a line with A and B in the standard form, separated by one space.
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.
Sample Input:
3.2.1 10.16.27
Sample Output:
14.1.28
题目分析:进制转化
#define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
long long N1[];
long long N2[];
long long Flag[];
long long C[] = { ,, };
int main()
{
scanf("%lld.%lld.%lld %lld.%lld.%lld", &N1[], &N1[], &N1[], &N2[], &N2[], &N2[]);
for (long long i = ; i >=; i--)
{
N1[i] += N2[i] + Flag[i];
if (N1[i] >=C[i]&&i!=)
{
Flag[i - ] = N1[i]/C[i];
N1[i] %= C[i];
}
}
printf("%lld.%lld.%lld", N1[], N1[], N1[]);
}
1058 A+B in Hogwarts (20分)(水)的更多相关文章
- PAT 甲级 1058 A+B in Hogwarts (20 分) (简单题)
1058 A+B in Hogwarts (20 分) If you are a fan of Harry Potter, you would know the world of magic ha ...
- 1058 A+B in Hogwarts (20分)
1058 A+B in Hogwarts (20分) 题目: If you are a fan of Harry Potter, you would know the world of magic h ...
- PAT Advanced 1058 A+B in Hogwarts (20 分)
If you are a fan of Harry Potter, you would know the world of magic has its own currency system -- a ...
- 【PAT甲级】1058 A+B in Hogwarts (20 分)
题意: 输入两组,每组三个非负整数A,B,C(A<=1e7,B<17,C<29),输出相加的和.(类似个位上29进制,十位上17进制运算) AAAAAccepted code: #d ...
- PAT甲题题解-1058. A+B in Hogwarts (20)-大水题
无语,这种水题还出,浪费时间,但又不得不A... #include <iostream> #include <cstdio> #include <algorithm> ...
- 1042 Shuffling Machine (20分)(水)
Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techn ...
- 1035 Password (20分)(水)
To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem ...
- 1058 A+B in Hogwarts (20)
#include <stdio.h> int main() { ]; ]; ],&ans1[],&ans1[],&ans2[],&ans2[],&a ...
- PAT (Advanced Level) 1058. A+B in Hogwarts (20)
简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...
随机推荐
- 【Oracle】RAC的多实例数据迁移至单机的多实例。
思路:一般的思路可以通过RMAN进行数据的恢复.由于数据库可以停机,因此,这次试用数据泵(expdp,impdp)进行数据 的导入导出. 1.源数据库导出 通过编写导出shell脚本导出数据,如下: ...
- 搭建OpenStack私有云准备工作
Centos7安装完成后克隆其他子节点 首先在VMware中:右击 虚拟机controller-->设置-->添加-->网络适配器,然后做如下设置: 在VMware中操作 点击:克隆 ...
- vue基础----key的作用
1.key 在虚拟dom中 标记不同的结构,for循环中需要添加上 <body> <div id="app"> <div v-if="fla ...
- juery 实现选项卡
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...
- tkinter学习1
GUI 用户交互界面 tkinter 介绍 tkinter是 python自带的gui库,对图像处理库tk的封装 #导入tkinter库 import tkinter #创建主窗口对象 root = ...
- (转)浅析epoll – epoll例子以及分析
原文地址:http://www.cppfans.org/1419.html 浅析epoll – epoll例子以及分析 上篇我们讲到epoll的函数和性能.这一篇用用这些个函数,给出一个最简单的epo ...
- golang socket编程 net.Conn IO.EOF解读
结论 首先,先定义下我的理解,当在Read时,收到一个IO.EOF,代表的就是对端已经关闭了发送的通道,通常来说是发起了FIN. 那么根据自己的实际业务,就可以进行判断,这里的IO.EOF到底该怎么利 ...
- 理解Golang组件protobuf
什么是protobuf protocol buffers 是一种语言无关.平台无关.可扩展的序列化结构数据的方法,它可用于(数据)通信协议.数据存储等.是一种灵活,高效,自动化机制的结构数据序列化方法 ...
- 物联网 软硬件系统 树莓派 单片机 esp32 小程序 网页 开发 欢迎相互交流学习~
物联网软硬件开发 知识分享 多年学生项目开发经验 物联网 软硬件系统 树莓派 单片机 esp32 小程序 网页 开发 欢迎相互交流学习~ http://39.105.218.125:9000/
- Django模型层ORM学习笔记
一. 铺垫 1. 连接Django自带数据库sqlite3 之前提到过Django自带一个叫做sqlite3的小型数据库,当我们做本地测试时,可以直接在sqlite3上测试.不过该数据库是小型的,在有 ...