B. Multithreading

Emuskald is addicted to Codeforces, and keeps refreshing the main page not to miss any changes in the "recent actions" list. He likes to read thread conversations where each thread consists of multiple messages.

Recent actions shows a list of n different threads ordered by the time of the latest message in the thread. When a new message is posted in a thread that thread jumps on the top of the list. No two messages of different threads are ever posted at the same time.

Emuskald has just finished reading all his opened threads and refreshes the main page for some more messages to feed his addiction. He notices that no new threads have appeared in the list and at the i-th place in the list there is a thread that was at the ai-th place before the refresh. He doesn't want to waste any time reading old messages so he wants to open only threads with new messages.

Help Emuskald find out the number of threads that surely have new messages. A thread x surely has a new message if there is no such sequence of thread updates (posting messages) that both conditions hold:

  1. thread x is not updated (it has no new messages);
  2. the list order 1, 2, ..., n changes to a1, a2, ..., an.

Input

The first line of input contains an integer n, the number of threads (1 ≤ n ≤ 105). The next line contains a list of n space-separated integers a1, a2, ..., an where ai (1 ≤ ai ≤ n) is the old position of the i-th thread in the new list. It is guaranteed that all of the ai are distinct.

Output

Output a single integer — the number of threads that surely contain a new message.

Examples

input

5
5 2 1 3 4

output

2

input

3
1 2 3

output

0

input

4
4 3 2 1

output

3

Note

In the first test case, threads 2 and 5 are placed before the thread 1, so these threads must contain new messages. Threads 1, 3 and 4 may contain no new messages, if only threads 2 and 5 have new messages.

In the second test case, there may be no new messages at all, since the thread order hasn't changed.

In the third test case, only thread 1 can contain no new messages.

水~~

#include<iostream>
using namespace std;
int a[100005];
int main(){
int n;
cin>>n;
int sum=0;
for(int i=1;i<=n;i++)
{
cin>>a[i]; }
for(int i=1;i<=n-1;i++)
if(a[i]>a[i+1])sum=i;
cout<<sum<<endl; }

codeforce 270B Multithreading的更多相关文章

  1. [.net 面向对象程序设计进阶] (18) 多线程(Multithreading)(三) 利用多线程提高程序性能(下)

    [.net 面向对象程序设计进阶] (18) 多线程(Multithreading)(二) 利用多线程提高程序性能(下) 本节导读: 上节说了线程同步中使用线程锁和线程通知的方式来处理资源共享问题,这 ...

  2. [.net 面向对象程序设计进阶] (17) 多线程(Multithreading)(二) 利用多线程提高程序性能(中)

    [.net 面向对象程序设计进阶] (17) 多线程(Multithreading)(二) 利用多线程提高程序性能(中) 本节要点: 上节介绍了多线程的基本使用方法和基本应用示例,本节深入介绍.NET ...

  3. [.net 面向对象程序设计进阶] (16) 多线程(Multithreading)(一) 利用多线程提高程序性能(上)

    [.net 面向对象程序设计进阶] (16) 多线程(Multithreading)(一) 利用多线程提高程序性能(上) 本节导读: 随着硬件和网络的高速发展,为多线程(Multithreading) ...

  4. Implicit and Explicit Multithreading MULTITHREADING AND CHIP MULTIPROCESSORS

    COMPUTER ORGANIZATION AND ARCHITECTURE DESIGNING FOR PERFORMANCE NINTH EDITION The concept of thread ...

  5. MULTITHREADING AND CHIP MULTIPROCESSORS

    COMPUTER ORGANIZATION AND ARCHITECTURE DESIGNING FOR PERFORMANCE NINTH EDITION The most important me ...

  6. Codeforce - Street Lamps

    Bahosain is walking in a street of N blocks. Each block is either empty or has one lamp. If there is ...

  7. Codeforce Round #216 Div2

    e,还是写一下这次的codeforce吧...庆祝这个月的开始,看自己有能,b到什么样! cf的第二题,脑抽的交了错两次后过了pretest然后system的挂了..脑子里还有自己要挂的感觉,果然回头 ...

  8. Multithreading annd Grand Central Dispatch on ios for Beginners Tutorial-多线程和GCD的入门教程

    原文链接:Multithreading and Grand Central Dispatch on iOS for Beginners Tutorial Have you ever written a ...

  9. Part 86 to 88 Talking about Multithreading in C#

    Part 86   Multithreading in C# What is a Process: Process is what the operating system uses to facil ...

随机推荐

  1. Maybatis的一些总结(三:增删改查)

    回顾一个点 之前不懂这句: UserMapper userMapper = sqlSession.getMapper(UserMapper.class); 现在理解了一点点,相当于实现了userMap ...

  2. 使用rem配置PC端自适应大屏

    效果如下 使得大屏不论在什么宽高比例依然能展示全部数据 安装 npm install -S postcss-pxtorem rem配置思路 原先的rem函数是能解决大部分的问题的,如果展示不全,也可以 ...

  3. Java成长第五集--面向对象设计的五大原则

    S.O.L.I.D 是面向对象设计(OOD)和面向对象编程(OOP)中的几个重要编码原则(Programming Priciple)的首字母缩写.以下图说明: 下面就个人的理解来说说这五大原则的含义到 ...

  4. 一站式WebAPI与认证授权服务

    保护WEBAPI有哪些方法? 微软官方文档推荐了好几个: Azure Active Directory Azure Active Directory B2C (Azure AD B2C)] Ident ...

  5. linux下DNS服务器搭建,正反向解析配置

    dns服务器之前自己搭建玩过,一段时间不搞,加上当时没写文档,基本忘光光了,这次老实了,写个文档记下来,方便以后查阅. 1.服务器准备 为了避免不必要的问题,关闭防火墙,关闭selinux,hosts ...

  6. C. Beautiful Regional Contest

    用前缀和写一直wa.. 思路:让金牌和银牌最少,通多增加铜牌的方式来扩大总奖牌的个数. #include<bits/stdc++.h> using namespace std; map&l ...

  7. F - Bone Collector

    Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like ...

  8. 批量重命名脚本(Python)

    便携的批处理脚本,代码如下: import os import sys def rename(): path=input("请输入路径(例如D:/picture):") name= ...

  9. Linux学习笔记(七)关机、重启及常用的网络命令

    关机.重启命令 sync shutdown reboot init sync 英文原意:flush file system buffers 功能:刷新文件系统缓冲区,将内存中的数据保存到硬盘中 语法: ...

  10. SpringCloud(七)超时、重试

    一.Ribbon(单独配置) 可以通过ribbon.xx来进行全局配置.也可以通过服务名.ribbon.xx来对指定服务配置 全局配置: ribbon: ConnectTimeout: 3000 #连 ...