Sanatorium

CodeForces - 732C

Vasiliy spent his vacation in a sanatorium, came back and found that he completely forgot details of his vacation!

Every day there was a breakfast, a dinner and a supper in a dining room of the sanatorium (of course, in this order). The only thing that Vasiliy has now is a card from the dining room contaning notes how many times he had a breakfast, a dinner and a supper (thus, the card contains three integers). Vasiliy could sometimes have missed some meal, for example, he could have had a breakfast and a supper, but a dinner, or, probably, at some days he haven't been at the dining room at all.

Vasiliy doesn't remember what was the time of the day when he arrived to sanatorium (before breakfast, before dinner, before supper or after supper), and the time when he left it (before breakfast, before dinner, before supper or after supper). So he considers any of these options. After Vasiliy arrived to the sanatorium, he was there all the time until he left. Please note, that it's possible that Vasiliy left the sanatorium on the same day he arrived.

According to the notes in the card, help Vasiliy determine the minimum number of meals in the dining room that he could have missed. We shouldn't count as missed meals on the arrival day before Vasiliy's arrival and meals on the departure day after he left.

Input

The only line contains three integers bd and s (0 ≤ b, d, s ≤ 1018,  b + d + s ≥ 1) — the number of breakfasts, dinners and suppers which Vasiliy had during his vacation in the sanatorium.

Output

Print single integer — the minimum possible number of meals which Vasiliy could have missed during his vacation.

Examples

Input
3 2 1
Output
1
Input
1 0 0
Output
0
Input
1 1 1
Output
0
Input
1000000000000000000 0 1000000000000000000
Output
999999999999999999

Note

In the first sample, Vasiliy could have missed one supper, for example, in case he have arrived before breakfast, have been in the sanatorium for two days (including the day of arrival) and then have left after breakfast on the third day.

In the second sample, Vasiliy could have arrived before breakfast, have had it, and immediately have left the sanatorium, not missing any meal.

In the third sample, Vasiliy could have been in the sanatorium for one day, not missing any meal.

sol:小学奥数题,找到最多的那个,一个个减过去即可

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
ll s=;
bool f=;
char ch=' ';
while(!isdigit(ch))
{
f|=(ch=='-'); ch=getchar();
}
while(isdigit(ch))
{
s=(s<<)+(s<<)+(ch^); ch=getchar();
}
return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
if(x<)
{
putchar('-'); x=-x;
}
if(x<)
{
putchar(x+''); return;
}
write(x/);
putchar((x%)+'');
return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
ll a,b,c;
int main()
{
R(a) ;R(b); R(c);
ll Max=max(max(a,b),c),ans=;
if(a<Max) ans+=Max-a-;
if(b<Max) ans+=Max-b-;
if(c<Max) ans+=Max-c-;
Wl(ans);
return ;
}
/*
input
3 2 1
output
1 Input
1 0 0
Output
0 Input
1 1 1
Output
0 Input
1000000000000000000 0 1000000000000000000
Output
999999999999999999
*/

codeforces732C的更多相关文章

随机推荐

  1. redis学习(五)——Set数据类型

    一.概述 在Redis中,我们可以将Set类型看作为没有排序的字符集合,和List类型一样,我们也可以在该类型的数据值上执行添加.删除或判断某一元素是否存在等操作.需要说明的是,这些操作的时间复杂度为 ...

  2. IntelliJ IDEA(一) :安装与破解

    前言 我是从eclipse转IDEA的,对于习惯了eclipse快捷键的我来说,转IDEA开始很不习惯,IDEA快捷键多,组合多,记不住,虽然可以设置使用eclipse的快捷键,但是总感觉怪怪的.开始 ...

  3. Java获取文件Content-Type的四种方法

    HTTP Content-Type在线工具 有时候我们需要获取本地文件的Content-Type,已知 Jdk 自带了三种方式来获取文件类型. 另外还有第三方包 Magic 也提供了API.Magic ...

  4. 史上最全面的Neo4j使用指南

    Neo4j图形数据库教程 Neo4j图形数据库教程 第一章:介绍 Neo4j是什么 Neo4j的特点 Neo4j的优点 第二章:安装 1.环境 2.下载 3.开启远程访问 4.测试 第三章:CQL 1 ...

  5. 如何选择分布式事务形态(TCC,SAGA,2PC,补偿,基于消息最终一致性等等)

    各种形态的分布式事务 分布式事务有多种主流形态,包括: 基于消息实现的分布式事务 基于补偿实现的分布式事务(gts/fescar自动补偿的形式) 基于TCC实现的分布式事务 基于SAGA实现的分布式事 ...

  6. Webpack+Typescript 简易配置

    教程:https://www.cnblogs.com/yasepix/p/9294499.html http://developer.egret.com/cn/github/egret-docs/ex ...

  7. GitHub上README.md编写教程(基本语法)

    一.标题写法: 第一种方法: 1.在文本下面加上 等于号 = ,那么上方的文本就变成了大标题.等于号的个数无限制,但一定要大于0个哦.. 2.在文本下面加上 下划线 - ,那么上方的文本就变成了中标题 ...

  8. Eclipse中Git的使用以及IDEA中Git的使用

    一.Eclipse中Git解决冲突步骤: 1.进行文件对比,将所有的文件添加到序列. 2.commit文件到本地仓库. 3.pull将远程仓库的代码更新到本地,若有冲突则会所有的文件显示冲突状态(真正 ...

  9. haoop笔记

    : //:什么是hadoop? hadoop是解决大数据问题的一整套技术方案 :hadoop的组成? 核心框架 分布式文件系统 分布式计算框架 分布式资源分配框架 hadoop对象存储 机器计算 :h ...

  10. CentOS6.5配置 cron

    CentOS6.5配置 cron 任务 - mengjiaoduan的博客 - CSDN博客https://blog.csdn.net/mengjiaoduan/article/details/649 ...