A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all n squad
soldiers to line up on the parade ground.

By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the
soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose
height is maximum or minimum. Only the heights of the first and the last soldier are important.

For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence(4,
3, 1, 2, 2) wrong.

Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct.

Input

The first input line contains the only integer n (2 ≤ n ≤ 100)
which represents the number of soldiers in the line. The second line contains integers a1, a2, ..., an (1 ≤ ai ≤ 100)
the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers a1, a2, ..., an are
not necessarily different.

Output

Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like.

Sample test(s)
input
4
33 44 11 22
output
2
input
7
10 10 58 31 63 40 76
output
10

相当于一个简单的冒泡排序了,只是不用直接排序,仅仅是计算一下而已。

注意

1 最大值和最小值交换的时候能够降低一次交换的。

2 元素是会反复的。

#include <iostream>
using namespace std; namespace{
static const int MAX_VAL = (int) 1E9;
static const int MIN_VAL = (int) 1E-9;
}
void ArrivaloftheGeneral()
{
int n, max_i, min_i, max_n = MIN_VAL, min_n = MAX_VAL, a;
cin>>n;
for (unsigned i = 0; i < n; i++)
{
cin>>a;
if (a <= min_n)
{
min_n = a;
min_i = i;
}
if (a > max_n)
{
max_n = a;
max_i = i;
}
}
cout<<max_i + (n - min_i - 1) - (max_i > min_i);
}

codeforces Arrival of the General 题解的更多相关文章

  1. Codeforces Round #182 (Div. 1)题解【ABCD】

    Codeforces Round #182 (Div. 1)题解 A题:Yaroslav and Sequence1 题意: 给你\(2*n+1\)个元素,你每次可以进行无数种操作,每次操作必须选择其 ...

  2. Codeforces Round #608 (Div. 2) 题解

    目录 Codeforces Round #608 (Div. 2) 题解 前言 A. Suits 题意 做法 程序 B. Blocks 题意 做法 程序 C. Shawarma Tent 题意 做法 ...

  3. Codeforces Round #525 (Div. 2)题解

    Codeforces Round #525 (Div. 2)题解 题解 CF1088A [Ehab and another construction problem] 依据题意枚举即可 # inclu ...

  4. Codeforces Round #528 (Div. 2)题解

    Codeforces Round #528 (Div. 2)题解 A. Right-Left Cipher 很明显这道题按题意逆序解码即可 Code: # include <bits/stdc+ ...

  5. Codeforces Round #466 (Div. 2) 题解940A 940B 940C 940D 940E 940F

    Codeforces Round #466 (Div. 2) 题解 A.Points on the line 题目大意: 给你一个数列,定义数列的权值为最大值减去最小值,问最少删除几个数,使得数列的权 ...

  6. Codeforces Round #677 (Div. 3) 题解

    Codeforces Round #677 (Div. 3) 题解 A. Boring Apartments 题目 题解 简单签到题,直接数,小于这个数的\(+10\). 代码 #include &l ...

  7. Codeforces Round #665 (Div. 2) 题解

    Codeforces Round #665 (Div. 2) 题解 写得有点晚了,估计都官方题解看完切掉了,没人看我的了qaq. 目录 Codeforces Round #665 (Div. 2) 题 ...

  8. Codeforces Round #160 (Div. 1) 题解【ABCD】

    Codeforces Round #160 (Div. 1) A - Maxim and Discounts 题意 给你n个折扣,m个物品,每个折扣都可以使用无限次,每次你使用第i个折扣的时候,你必须 ...

  9. Codeforces Round #383 (Div. 2) 题解【ABCDE】

    Codeforces Round #383 (Div. 2) A. Arpa's hard exam and Mehrdad's naive cheat 题意 求1378^n mod 10 题解 直接 ...

随机推荐

  1. 富文本编辑器CKEDITOR的使用配置(问题注解)

    CKEDITOR是一款非常轻便的富文本编辑器,如上图:参考网上的使用方法,我以.net为例,在aspx页面使用, 准备工作:首先要下载控件包,将解压后的整个文件夹添加到项目根目录. 第一步:引用js, ...

  2. DIV布局之道二:DIV块的嵌套,DIV盒子模型

    本文讲解DIV块布局的第二种使用方式:嵌套.“DIV嵌套”在有些文献中也被称为“盒子模型”,说的通俗一点就是嵌套(一个大的DIV块内部又包含一个或多个DIV块). 请看如下代码: CSS部分: CSS ...

  3. Android Studio 打开弹出警告框

    1.Android Studio打开后,自己的项目没有打开,就弹出了警告框,重启之后依然弹出警告框: 警告框内容:"Cannot load project: java.lang.Illega ...

  4. 有关Ajax实现的两种方法

    首先我们来常见的Jquery式的Ajax写法,以及在java后台的取值 /** *AJAX */ function showLastTime(){ var facilityId = $('*[name ...

  5. Java中的HashMap和Hashtable

    代码: import java.util.*; public class test{ public static void main(String[] args) { HashMap hm = new ...

  6. FlowPlayer 参数说明

    <html xmlns="http://www.w3.org/1999/xhtml" > <head runat="server"> & ...

  7. HTML5 canvas 中的线条样式

    线条样式属性 lineCap        设置或返回线条的结束端点样式 butt         默认.向线条的每个末端添加平直的边缘. round         向线条的每个末端添加圆形线帽. ...

  8. QQ原创表情添加

    有时候与QQ好友聊天时会收到一些自己比较喜欢的原创表情,如果我们想把这些QQ不支持直接保存的原创表情保存到电脑上该怎么做呢?我们以原创表情图片为例简单介绍一下. 首先,先建立一个存放表情图片的文件夹. ...

  9. 手机端禁止iPhone字体放大

    /*禁止iphone字体放大 */ html { -webkit-text-size-adjust: none; }

  10. jquery radio取值,checkbox取值,select取值,radio选中,checkbox选中,select选中

    jQuery获取Select选择的Text和Value: 语法解释: 1. $("#select_id").change(function(){//code...}); //为Se ...