2019-05-30

20:19:57

加油!!!

sort(a + 1, a + 5);

卡了一会儿

 #include <bits/stdc++.h>
using namespace std;
typedef long long ll; int main()
{
ll a[], e;
ll sum = ;
for (int i = ; i <= ; i++)
{
scanf("%lld ", &a[i]);
}
scanf("%lld", &e);
if (a[] < a[] || a[] < a[])
{
cout << ;
return ;
}
else
{
sort(a + , a + );
sum = a[] - a[] + ;
if (e >= a[] && e <= a[])
{
sum--;
}
cout << sum << endl;
} return ;
}

CodeForces A. Meeting of Old Friends的更多相关文章

  1. Codeforces 714A Meeting of Old Friends

    A. Meeting of Old Friends time limit per test:1 second memory limit per test:256 megabytes input:sta ...

  2. CodeForces 144B Meeting

    暴力. 题目只要求计算边上的点就可以了,一开始没看清题意,把内部的也算进去了.内部的计算可以延迟标记一下,但这题没有必要. #include<map> #include<set> ...

  3. codeforces 782B The Meeting Place Cannot Be Changed (三分)

    The Meeting Place Cannot Be Changed Problem Description The main road in Bytecity is a straight line ...

  4. Codeforces Round #375 (Div. 2) A. The New Year: Meeting Friends 水题

    A. The New Year: Meeting Friends 题目连接: http://codeforces.com/contest/723/problem/A Description There ...

  5. Codeforces Round #371 (Div. 2) A. Meeting of Old Friends 水题

    A. Meeting of Old Friends 题目连接: http://codeforces.com/contest/714/problem/A Description Today an out ...

  6. Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) B. The Meeting Place Cannot Be Changed

    地址:http://codeforces.com/contest/782/problem/B 题目: B. The Meeting Place Cannot Be Changed time limit ...

  7. Jury Meeting CodeForces - 854D

    Jury Meeting CodeForces - 854D 思路:暴力枚举会议开始的那一天(只需用所有向0点飞的航班的那一天+1去枚举即可),并计算所有人此情况下去0点和从0点出来的最小花费. 具体 ...

  8. Codeforces 420 B. Online Meeting

    B. Online Meeting time limit per test 1 second memory limit per test 256 megabytes input standard in ...

  9. Jury Meeting CodeForces - 854D (前缀和维护)

    Country of Metropolia is holding Olympiad of Metrpolises soon. It mean that all jury members of the ...

随机推荐

  1. GCD & Operation queues & Thread

    One of the technologies for starting tasks asynchronously is Grand Central Dispatch (GCD). This tech ...

  2. tidyverse生态链

    一套完整的数据分析流程 , 如下图所示 从图中可以看到,整个流程包括读取数据,整洁数据,数据探索和交流部分.经过前两部分, 我们可以得到一个整理好的数据,它的每一行都是一个样本 , 每一列是一个变量. ...

  3. 【原】PHPExcel导出Excel

    1.引入相关公共库PHPExcel 2.编写公共函数 public function exportExcel($excelTitle,$data,$filename='',$column_width= ...

  4. Jmeter读取excel表中用例数据实现接口压测

    传统的接口测试,都是在接口中手动输入不同用例准备的多种场景参数数据,一遍一遍的输入来执行多个不同的用例,但是现在利用excel表格准备各种类型的数据,使用Jmeter中Jmeter CSV Data ...

  5. cogs——66. [HAOI2004模拟] 数列问题

    66. [HAOI2004模拟] 数列问题 本以为会TLE,可... dfs水题(很基础) #include<bits/stdc++.h> using namespace std; ],a ...

  6. uva 1585 Score(Uva-1585)

    vj:https://vjudge.net/problem/UVA-1585 不多说水题一个o一直加x就加的变为0 我的代码 #include <iostream> #include &l ...

  7. 腾讯云,搭建 Discuz 个人论坛

    准备 LAMP 环境 任务时间:30min ~ 60min LAMP 是 Linux.Apache.MySQL 和 PHP 的缩写,是 Discuz 论坛系统依赖的基础运行环境.我们先来准备 LAMP ...

  8. PAT 1100. Mars Numbers

    People on Mars count their numbers with base 13: Zero on Earth is called "tret" on Mars. T ...

  9. docker的容器可视化工具portainer

    1.搜索镜像 [root@holly ~]# docker search portainer 2.下载portainer [root@holly ~]# docker pull portainer/p ...

  10. ACM的你伤不起

    劳资六年前开始搞ACM啊!!!!!!!!!!  从此踏上了尼玛不归路啊!!!!!!!!!!!!  谁特么跟劳资讲算法是程序设计的核心啊!!!!!!  尼玛除了面试题就没见过用算法的地方啊!!!!!!  ...