B. Minimize the Permutation

You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

You can perform at most n−1 operations with the given permutation (it is possible that you don't perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.

Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.

You can see the definition of the lexicographical order in the notes section.

You have to answer q independent test cases.

For example, let's consider the permutation [5,4,1,3,2]. The minimum possible permutation we can obtain is [1,5,2,4,3] and we can do it in the following way:

perform the second operation (swap the second and the third elements) and obtain the permutation [5,1,4,3,2];

perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5,1,4,2,3];

perform the third operation (swap the third and the fourth elements) and obtain the permutation [5,1,2,4,3].

perform the first operation (swap the first and the second elements) and obtain the permutation [1,5,2,4,3];

Another example is [1,2,4,3]. The minimum possible permutation we can obtain is [1,2,3,4] by performing the third operation (swap the third and the fourth elements).

Input

The first line of the input contains one integer q (1≤q≤100) — the number of test cases. Then q test cases follow.

The first line of the test case contains one integer n (1≤n≤100) — the number of elements in the permutation.

The second line of the test case contains n distinct integers from 1 to n — the given permutation.

Output

For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.

Example

input

4

5

5 4 1 3 2

4

1 2 4 3

1

1

4

4 3 2 1

output

1 5 2 4 3

1 2 3 4

1

1 4 3 2

Note

Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i≤n that for all j from 1 to i−1 the condition pj=qj is satisfied, and pi<qi

p=[1,3,5,2,4] is less than q=[1,3,5,4,2] (such i=4 exists, that pi<qi and for each j<i holds pj=qj),

p=[1,2] is less than q=[2,1] (such i=1 exists, that pi<qi and for each j<i holds pj=qj).

题意

q次询问,每次询问给你长度为n的排列,然后你每次可以选择一个位置i和i+1的数字进行交换。但是每个位置只能交换一次,问你反转若干次后,这个排列最小是多少?

题解

贪心,每次选择最小的数往前走就好了。

代码

#include<bits/stdc++.h>
using namespace std; vector<int>p;
int pos[105];
int vis[105];
void solve(){
p.clear();
int n;
scanf("%d",&n);
memset(vis,0,sizeof(vis));
memset(pos,0,sizeof(pos));
for(int i=0;i<n;i++){
int x;scanf("%d",&x);
p.push_back(x);
pos[x]=i;
}
for(int i=1;i<=n;i++){
int flag = 1;
while(flag==1){
if(pos[i]>0&&vis[pos[i]-1]==0){
vis[pos[i]-1]=1;
int now=pos[i],pnow=pos[i]-1;
swap(p[now],p[pnow]);
swap(pos[p[now]],pos[p[pnow]]);
}else{
flag=0;
}
}
vis[pos[i]]=1;
//for(int i=0;i<p.size();i++){
// cout<<vis[i]<<" ";
//}
//cout<<endl;
}
for(int i=0;i<p.size();i++){
cout<<p[i]<<" ";
}
cout<<endl; }
int main(){
int t;
scanf("%d",&t);
while(t--)solve();
}

Codeforces Round #598 (Div. 3) B. Minimize the Permutation 贪心的更多相关文章

  1. Codeforces Round #598 (Div. 3) B Minimize the Permutation

    B. Minimize the Permutation You are given a permutation of length nn. Recall that the permutation is ...

  2. Codeforces Round #598 (Div. 3) D. Binary String Minimizing 贪心

    D. Binary String Minimizing You are given a binary string of length n (i. e. a string consisting of ...

  3. Codeforces Round #598 (Div. 3)- E. Yet Another Division Into Teams - 动态规划

    Codeforces Round #598 (Div. 3)- E. Yet Another Division Into Teams - 动态规划 [Problem Description] 给你\( ...

  4. Codeforces Round #297 (Div. 2)C. Ilya and Sticks 贪心

    Codeforces Round #297 (Div. 2)C. Ilya and Sticks Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  ...

  5. 【CF1256】Codeforces Round #598 (Div. 3) 【思维+贪心+DP】

    https://codeforces.com/contest/1256 A:Payment Without Change[思维] 题意:给你a个价值n的物品和b个价值1的物品,问是否存在取物方案使得价 ...

  6. Codeforces Round #598 (Div. 3)

    传送门 A. Payment Without Change 签到. Code /* * Author: heyuhhh * Created Time: 2019/11/4 21:19:19 */ #i ...

  7. Codeforces Round #598 (Div. 3) A,B,C,D{E,F待补}

    A. Payment Without Change   #include<bits/stdc++.h> using namespace std; #define int long long ...

  8. Codeforces Round #598 (Div. 3) E. Yet Another Division Into Teams dp

    E. Yet Another Division Into Teams There are n students at your university. The programming skill of ...

  9. Codeforces Round #598 (Div. 3)E(dp路径转移)

    题:https://codeforces.com/contest/1256/problem/E 题意:给一些值,代表队员的能力值,每组要分3个或3个以上的人,然后有个评价值x=(队里最大值-最小值), ...

随机推荐

  1. arcgis api for javascript 学习(一) 调用在线发布的动态地图

    1.图中显示为arcgis软件中显示的地图文件 2.调用动态地图主要的是知道动态地图的URL地址 3.通过IDE(webstorm)调用动态地图,如图 4.话不多说,直接上代码 <!DOCTYP ...

  2. 分析Android APK-反编译修改打包

    2.2 这个章节的主要作用就是,修改一个别人的app,在里边增加一段自己的广告代码. 2.2.1 UAA 编译修改,工具安装配置之前讲过,无需再赘述.我们找到了一款游戏app的apk, 找到所在的ap ...

  3. Android studio 3.4 新建项目报错Error:unable to resolve dependency for app@。。。解决办法

    试过网上很多的例子,有的设置Go to `File->Settings->Build, Execution, Deployment->Gradle->Uncheck Offli ...

  4. Mysql模式匹配两种方法

    一.使用LIKE或NOT LIKE比较操作符 使用 "_" 匹配任何单个字符,而 "%" 匹配任意数量的字符(包括零字符): 例如: 1.要想找出以“b”开头的 ...

  5. 小程序--log居中 失焦获取表单中的值

    value="{{username}}" 绑定值 值在js文件的data中 pa==>Vant-Weap中表单中的值,不是双向绑定的. 你获取值后, 值并没有在对用的data ...

  6. GUI程序分析实例

    GUI程序开发概述 GUI程序开发原理 GetMessage(&msg)将消息队列中的消息取出来,在循环中进行处理. GUI程序开发的本质

  7. .net core 在 Docker 上的部署

    Docker可以说是现在微服务,DevOps的基础,咱们.Net Core自然也得上Docker..Net Core发布到Docker容器的教程网上也有不少,但是今天还是想来写一写.你搜.Net co ...

  8. IT兄弟连 HTML5教程 CSS3揭秘 CSS常见的样式属性和值2

    3  背景属性 大多数HTML元素都允许控制背景,包括背景颜色.背景图像.背景重复.背景附件.背景位置等属性.常见的控制背景属性.值及描述如表2所示. 表2  CSS中常见的控制背景的属性 除了使用表 ...

  9. 基于appium的fixture应用之代码重构

    一.痛点分析 在appium自动化中,会话启动参数较多,我们使用了yaml配置文件来进行管理,并使用了PyYaml模块进行yaml文件内容的读取,我们知道,在测试场景中,不可能只会用到一种启动类型的参 ...

  10. Oracle - SPM固定执行计划(二)

    一.前言 前面文章(https://www.cnblogs.com/ddzj01/p/11365541.html)给大家介绍了当一条sql有多个执行计划时,如何通过spm去绑定其中一条执行计划.本文将 ...