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
- 【python pip】一招解决pip下载过慢问题
目录 概述 壹:问题描述 贰:解决过程 一.问题分析 二.问题解决 方法一:下载时加入参数-i [镜像源地址] 方法二:设置源 三.国内镜像源地址 叁:作者有话 作者 概述 在我们经常使用pip安装插 ...
- deepin15.11安装N卡驱动,实测!!!(可解决N卡电脑关机卡屏)
前言:deepin(深度)是一款由武汉深之度公司研发的一款适合国人日常学习的linux系统,其UI精美,美过Mac.它对于中国用户的一个亮点就是QQ微信等国软件傻瓜式安装(类似安卓应用商店安装),如果 ...
- php实现下载功能
<?php header("Content-type:text/html;charset=utf-8"); $file_name="1.text"; // ...
- xpath提取标签和内容
转:https://segmentfault.com/q/1010000012110138/a-1020000012113020 <div> <table> <tr> ...
- Spark入门(七)--Spark的intersection、subtract、union和distinc
Spark的intersection intersection顾名思义,他是指交叉的.当两个RDD进行intersection后,将保留两者共有的.因此对于RDD1.intersection(RDD2 ...
- IPv4地址表示法详解
在TCP/IP协议中,IP地址是一个最基本的概念,本文就来参考<计算机网络>谢希仁 这本书,总结一下IPv4地址表示法的发展阶段,做个读书笔记. IP地址的编址方法共经过了三个历史阶段: ...
- HTTP请求中Get和Post请求的区别?
分类 Get的请求方式 1.直接在浏览器地址栏输入某个地址. 2.点击链接地址. 3.表单的默认提交方式或者设置为method="get". Post的请求方式 1.设置表单的me ...
- 原创】Java并发编程系列2:线程概念与基础操作
[原创]Java并发编程系列2:线程概念与基础操作 伟大的理想只有经过忘我的斗争和牺牲才能胜利实现. 本篇为[Dali王的技术博客]Java并发编程系列第二篇,讲讲有关线程的那些事儿.主要内容是如下这 ...
- Docker学习-私有仓库docker-registry的使用
1.从docker官方仓库下载registry 2.将registry放进容器内 3.在官方下载镜像上传本地仓库 4.私有仓库docker-registry使用的常见问题 5.配置阿里云镜像加速器 假 ...
- Android UI性能测试——使用 Gfxinfo 衡量性能
Android官方文档翻译 原文地址:https://developer.android.com/training/testing/performance参考:https://www.jianshu. ...