Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 275  Solved: 175
[Submit][Status][Discuss]

Description

Bessie the cow is a huge fan of card games, which is quite surprising, given her lack of opposable thumbs. Unfortunately, none of the other cows in the herd are good opponents. They are so bad, in fact, that they always play in a completely predictable fashion! Nonetheless, it can still be a challenge for Bessie to figure out how to win.
Bessie and her friend Elsie are currently playing a simple card game where they take a deck of 2N cards, conveniently numbered 1…2N, and divide them into N cards for Bessie and N cards for Elsie. The two then play N rounds, where in each round Bessie and Elsie both play a single card, and the player with the highest card earns a point.
Given that Bessie can predict the order in which Elsie will play her cards, please determine the maximum number of points Bessie can win.

奶牛Bessie和Elsie在玩一种卡牌游戏。一共有2N张卡牌,点数分别为1到2N,每头牛都会分到N张卡牌。

游戏一共分为N轮,因为Bessie太聪明了,她甚至可以预测出每回合Elsie会出什么牌。

每轮游戏里,两头牛分别出一张牌,点数大者获胜。

Bessie现在想知道,自己最多能获胜多少轮?

Input

The first line of input contains the value of N (1≤N≤50,000).
The next N lines contain the cards that Elsie will play in each of the successive rounds of the game. Note that it is easy to determine Bessie's cards from this information.

Output

Output a single line giving the maximum number of points Bessie can score.

Sample Input

3
1
6
4

Sample Output

2

HINT

Here, Bessie must have cards 2, 3, and 5 in her hand, and she can use these to win at most 2 points by saving the 5 until the end to beat Elsie's 4.

Problem credits: Austin Bannister and Brian Dean

Source

Silver鸣谢Claris提供译文

set似乎可以强搞,利用set自排序的特性,再按照田忌赛马原则(?)大雾

 #include<iostream>
#include<cstdio>
#include<set>
#include<iterator>
#include<algorithm>
using namespace std; int n,node=;
int a[];
set<int> S;
set<int>::iterator iter; int main()
{
scanf("%d",&n);
for(int i=;i<=*n;i++) S.insert(i);
for(int i=;i<n;i++)
{
scanf("%d",&a[i]);
S.erase(a[i]);
}
sort(a,a+n);
for(iter=S.begin();iter!=S.end();iter++)
if(*iter>a[node]) node++;
printf("%d",node);
return ;
}

最新文章

  1. Linux LVM学习总结&mdash;&mdash;放大LV容量
  2. mutation annovar
  3. glusterFS的常用命令 (转)
  4. 北京程序员 VS 硅谷程序员(转)
  5. popupwindow 与 输入法
  6. iOS 下拉刷新和加载更多 (OC\Swift)
  7. 拿nodejs快速搭建简单Oauth认证和restful API server攻略
  8. [wikioi]乌龟棋
  9. EF+WCF怎样更新数据?
  10. 轻量级的内部测试过程r \\ u0026研发团队
  11. HDU 2063:过山车(偶匹配,匈牙利算法)
  12. 解决Jenkins performance-plugin 解析jmeter jtl报告NumberFormatException &quot;timeStamp&quot;
  13. android EditText与TextView几个常用的属性
  14. RabbitMQ封装实战
  15. [JLOI 2012]树
  16. EF下lambda与linq查询&amp;&amp;扩展方法
  17. 通过 docker 来搭建 Jenkins
  18. java报错:The type java.lang.Object cannot be resolved. It is indirectly referenced from required .class files
  19. Python的浮点数损失精度问题
  20. 寒假训练——搜索 G - Xor-Paths

热门文章

  1. Decorator模式(装饰器模式)
  2. maven POM总结
  3. Clion下载安装使用教程(Win+MinGW)
  4. map系统学习
  5. Java继承改进
  6. springboot 学习笔记(八)
  7. 《Head First 设计模式》之装饰者模式——饮料加工
  8. strlen()与mb_strlen()的作用分别是什么
  9. Windows 10 取消桌面右键“图像属性”、“图像选项”
  10. 《Unity預計算即時GI》笔记:一、基本概念与一些设置