Sanatorium
1 second
256 megabytes
standard input
standard output
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.
The only line contains three integers b, d 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.
Print single integer — the minimum possible number of meals which Vasiliy could have missed during his vacation.
3 2 1
1
1 0 0
0
1 1 1
0
1000000000000000000 0 1000000000000000000
999999999999999999
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.
分析:贪心,哪个最大,就以哪个开始到,然后模拟;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3fLL
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
const int maxn=2e5+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
inline ll read()
{
ll x=;int f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,k,t;
ll a[],ans,ma;
int main()
{
int i,j;
rep(i,,)scanf("%lld",&a[i]),ma=max(ma,a[i]);
rep(i,,)if(a[i]<ma)ans+=ma--a[i];
printf("%lld\n",ans);
//system("Pause");
return ;
}
Sanatorium的更多相关文章
- Codeforces Round #377 (Div. 2) C. Sanatorium 水题
C. Sanatorium time limit per test 1 second memory limit per test 256 megabytes input standard input ...
- CodeForces 732C Sanatorium
C. Sanatorium time limit per test 1 second memory limit per test 256 megabytes input standard input ...
- 【43.26%】【codeforces 732C】Sanatorium
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- CodeForces 732C Sanatorium (if-else)
题意:某人去旅游,记录了一共吃了多少顿饭,早中晚,但是有可能缺少,问你最少缺少了多少顿. 析:每把三个数排序,然后再一一对比,肯定是以最大数为主,其他两个肯定是缺少了. 代码如下: #pragma c ...
- codeforces 732
A. Buy a Shovel time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- Codeforces Round #377 (Div. 2) A B C D 水/贪心/贪心/二分
A. Buy a Shovel time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- codeforces732C
Sanatorium CodeForces - 732C Vasiliy spent his vacation in a sanatorium, came back and found that he ...
- Codeforces Round #377 (Div. 2)A,B,C,D【二分】
PS:这一场真的是上分场,只要手速快就行.然而在自己做的时候不用翻译软件,看题非常吃力非常慢,还有给队友讲D题如何判断的时候又犯了一个毛病,一定要心平气和,比赛也要保证,不要用翻译软件做题: Code ...
- Codeforces Round #377 (Div. 2)部分题解A+B+C!
A. Buy a Shovel 题意是很好懂的,一件商品单价为k,但他身上只有10块的若干和一张r块的:求最少买几件使得不需要找零.只需枚举数量判断总价最后一位是否为0或r即可. #include&l ...
随机推荐
- html中的a标签的target属性的四个值的区别?
target属性规定了在何处打开超链接的文档. 如果在一个 <a> 标签内包含一个 target 属性,浏览器将会载入和显示用这个标签的 href 属性命名的.名称与这个目标吻合的框架或者 ...
- What is Flux?
Pluralsight - React and Flux for Angular Developers 1. An architectural concept. It a idea. 2. Not a ...
- C++ 内存分析-valgrind
valgrind包括了以下几个比较重要的模块:memcheck, cachegrind, callgrind, helgrind, drd, massif, dhat, sgcheck, bbv. 还 ...
- [ An Ac a Day ^_^ ] CodeForces 691F Couple Cover 花式暴力
Couple Cover Time Limit: 3000MS Memory Limit: 524288KB 64bit IO Format: %I64d & %I64u Descri ...
- web通知
<html> <head> <title>桌面通知</title> <meta name="description" cont ...
- mvc路由参数注解
routes.IgnoreRoute("{resource}.axd/{*pathInfo}"); //过滤掉禁止访问的路由 routes.MapRoute( name: &quo ...
- sql分页比较简单快捷的方法
SELECT TOP 显示数量* FROM 表 WHERE (主键id>(SELECT MAX(主键id) FROM(SELECT TOP 页码数*显示数量 主键id FROM 表 ORDER ...
- java的string.split()分割特殊字符时注意点
[1]单个符号作为分隔符 String address="上海|上海市|闵行区|吴中路"; String[] splitAddress=address.s ...
- D - 娜娜梦游仙境系列——村民的怪癖
D - 娜娜梦游仙境系列——村民的怪癖 Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Othe ...
- SQL函数学习(三):convert()函数
格式:CONVERT(data_type,expression[,style])说明:此样式一般在时间类型(datetime,smalldatetime)与字符串类型(nchar,nvarchar,c ...