B. Little Robber Girl's Zoo

题目连接:

http://www.codeforces.com/contest/686/problem/B

Description

Little Robber Girl likes to scare animals in her zoo for fun. She decided to arrange the animals in a row in the order of non-decreasing height. However, the animals were so scared that they couldn't stay in the right places.

The robber girl was angry at first, but then she decided to arrange the animals herself. She repeatedly names numbers l and r such that r - l + 1 is even. After that animals that occupy positions between l and r inclusively are rearranged as follows: the animal at position l swaps places with the animal at position l + 1, the animal l + 2 swaps with the animal l + 3, ..., finally, the animal at position r - 1 swaps with the animal r.

Help the robber girl to arrange the animals in the order of non-decreasing height. You should name at most 20 000 segments, since otherwise the robber girl will become bored and will start scaring the animals again.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — number of animals in the robber girl's zoo.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the height of the animal occupying the i-th place.

Output

Print the sequence of operations that will rearrange the animals by non-decreasing height.

The output should contain several lines, i-th of the lines should contain two space-separated integers li and ri (1 ≤ li < ri ≤ n) — descriptions of segments the robber girl should name. The segments should be described in the order the operations are performed.

The number of operations should not exceed 20 000.

If the animals are arranged correctly from the start, you are allowed to output nothing.

Sample Input

4

2 1 4 3

Sample Output

1 4

Hint

题意

给你n个数,你需要把它变成非递减的序列

你每次操作可以给定一个[L,R]区间,使得这个区间的数l和l+1交换,l+2和l+3交换。。。R和R-1交换

然后最多输出20000次,使得有序。

题解:

为什么这么麻烦呢?操作

我每次只操作两个数不就好了,然后就像冒泡排序那样去写就好了。。。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 105;
int a[maxn],n;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=n;i>=1;i--){
for(int j=1;j<i;j++){
if(a[j]>a[j+1]){
swap(a[j],a[j+1]);
cout<<j<<" "<<j+1<<endl;
}
}
}
}

Codeforces Round #359 (Div. 2) B. Little Robber Girl's Zoo 水题的更多相关文章

  1. Codeforces Round #356 (Div. 2)B. Bear and Finding Criminals(水题)

    B. Bear and Finding Criminals time limit per test 2 seconds memory limit per test 256 megabytes inpu ...

  2. Codeforces Round #343 (Div. 2) A. Far Relative’s Birthday Cake 水题

    A. Far Relative's Birthday Cake 题目连接: http://www.codeforces.com/contest/629/problem/A Description Do ...

  3. Codeforces Round #385 (Div. 2) A. Hongcow Learns the Cyclic Shift 水题

    A. Hongcow Learns the Cyclic Shift 题目连接: http://codeforces.com/contest/745/problem/A Description Hon ...

  4. Codeforces Round #396 (Div. 2) A. Mahmoud and Longest Uncommon Subsequence 水题

    A. Mahmoud and Longest Uncommon Subsequence 题目连接: http://codeforces.com/contest/766/problem/A Descri ...

  5. 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 ...

  6. Codeforces Round #258 (Div. 2) C. Predict Outcome of the Game 水题

    C. Predict Outcome of the Game 题目连接: http://codeforces.com/contest/451/problem/C Description There a ...

  7. Codeforces Round #358 (Div. 2) B. Alyona and Mex 水题

    B. Alyona and Mex 题目连接: http://www.codeforces.com/contest/682/problem/B Description Someone gave Aly ...

  8. 【打CF,学算法——二星级】Codeforces Round #313 (Div. 2) B. Gerald is into Art(水题)

    [CF简单介绍] 提交链接:http://codeforces.com/contest/560/problem/B 题面: B. Gerald is into Art time limit per t ...

  9. Codeforces Round #294 (Div. 2)A - A and B and Chess 水题

    A. A and B and Chess time limit per test 1 second memory limit per test 256 megabytes input standard ...

随机推荐

  1. 洛谷P1038神经网络

    传送门啦 一个拓扑排序的题,感觉题目好难懂... #include <iostream> #include <cstdio> #include <cstring> ...

  2. T-sql语句修改数据库逻辑名、数据库名、物理名(sql2000)

    --更改MSSQL数据库物理文件名Sql语句的写法 --注意:要在活动监视器里面确保没有进程连接你要改名的数据库!!!!!!!!!!!!!!!!!!!! -- Sql语句如下 USE master - ...

  3. Effective STL 学习笔记 Item 34: 了解哪些算法希望输入有序数据

    Effective STL 学习笔记 Item 34: 了解哪些算法希望输入有序数据 */--> div.org-src-container { font-size: 85%; font-fam ...

  4. peda的官方文档说明

    peda在github上的官方文档,摘抄过来,方便查阅. 安装 git clone https://github.com/longld/peda.git ~/peda echo "sourc ...

  5. dp入门题目

    本文文旨,如题... 转载请注明出处... HDOJ 1176 免费馅饼 http://acm.hdu.edu.cn/showproblem.php?pid=1176 类似数塔,从底往上推,每次都是从 ...

  6. Java模拟按键

    JDK自带了Robot类,此类用于为测试自动化.自运行演示程序和其他需要控制鼠标和键盘的应用程序生成本机系统输入事件.Robot 的主要目的是便于 Java 平台实现自动测试. 详情可查看jdk1.6 ...

  7. XeLaTeX下如何以原大小显示PNG

    在XeLaTeX里直接使用\includegraphics{test.png}这样的命令引入PNG,可能会发现图片直接被缩放到占满文档宽度,这是因为PNG这种bitmap类型的图片里通常不会带上met ...

  8. HBase(二)CentOS7.5搭建HBase1.2.6HA集群

    一.安装前提 1.HBase 依赖于 HDFS 做底层的数据存储 2.HBase 依赖于 MapReduce 做数据计算 3.HBase 依赖于 ZooKeeper 做服务协调 4.HBase源码是j ...

  9. Spark(三)RDD与广播变量、累加器

    一.RDD的概述 1.1 什么是RDD RDD(Resilient Distributed Dataset)叫做弹性分布式数据集,是Spark中最基本的数据抽象,它代表一个不可变.可分区.里面的元素可 ...

  10. Deepin Linux安装MySQL方法

    sudo apt-get install mysql-server apt-get install mysql-client sudo apt-get install libmysqlclient-d ...