According to Wikipedia:

Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there. It repeats until no input elements remain.

Heap sort divides its input into a sorted and an unsorted region, and it iteratively shrinks the unsorted region by extracting the largest element and moving that to the sorted region. it involves the use of a heap data structure rather than a linear-time search to find the maximum.

Now given the initial sequence of integers, together with a sequence which is a result of several iterations of some sorting method, can you tell which sorting method we are using?

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=100). Then in the next line, N integers are given as the initial sequence. The last line contains the partially sorted sequence of the N numbers. It is assumed that the target sequence is always ascending. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in the first line either "Insertion Sort" or "Heap Sort" to indicate the method used to obtain the partial result. Then run this method for one more iteration and output in the second line the resuling sequence. It is guaranteed that the answer is unique for each test case. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

10

3 1 2 8 7 5 9 4 6 0

1 2 3 7 8 5 9 4 6 0

Sample Output 1:

Insertion Sort

1 2 3 5 7 8 9 4 6 0

Sample Input 2:

10

3 1 2 8 7 5 9 4 6 0

6 4 5 1 0 3 2 7 8 9

Sample Output 2:

Heap Sort

5 4 3 1 0 2 6 7 8 9

#include<iostream>
using namespace std;
int main(){
int N; cin>>N;
int i,j,flag=1,tag=0;
int num0[N],num1[N];
for(i=0;i<N;i++)
cin>>num0[i];
for(i=0;i<N;i++)
cin>>num1[i];
for(i=0;i<N-1;i++)
if(num1[i]>num1[i+1]) break;
for(j=i+1;j<N;j++)
if(num1[j]!=num0[j]) flag=0;
if(flag==0){
cout<<"Heap Sort"<<endl;
for(j=1;j<N;j++)
if(num1[(j-1)/2]<num1[j]) break;
j=j-1;
int temp=num1[j];
num1[j]=num1[0];
num1[0]=temp;
j--;
int parent,child;
temp=num1[0];
for(parent=0;2*parent+1<=j;parent=child){
child=parent*2+1;
if(child<j&&num1[child+1]>num1[child])
child++;
if(num1[child]>temp)
num1[parent]=num1[child];
else
break;
}
num1[parent]=temp;
}
else{
cout<<"Insertion Sort"<<endl;
int temp=num1[i+1];
for(i=i+1;i>0;i--)
if(num1[i-1]>temp) num1[i]=num1[i-1];
else break;
num1[i]=temp;
}
for(int k=0;k<N;k++)
if(tag++==0) cout<<num1[k];
else cout<<" "<<num1[k];
return 0;
}

PAT 1098. Insertion or Heap Sort的更多相关文章

  1. PAT甲级1098. Insertion or Heap Sort

    PAT甲级1098. Insertion or Heap Sort 题意: 根据维基百科: 插入排序迭代,消耗一个输入元素每次重复,并增加排序的输出列表.在每次迭代中,插入排序从输入数据中删除一个元素 ...

  2. PAT甲级——1098 Insertion or Heap Sort (插入排序、堆排序)

    本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90941941 1098 Insertion or Heap So ...

  3. pat 甲级 1098. Insertion or Heap Sort (25)

    1098. Insertion or Heap Sort (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yu ...

  4. 1098 Insertion or Heap Sort——PAT甲级真题

    1098 Insertion or Heap Sort According to Wikipedia: Insertion sort iterates, consuming one input ele ...

  5. 1098 Insertion or Heap Sort

    1098 Insertion or Heap Sort (25 分) According to Wikipedia: Insertion sort iterates, consuming one in ...

  6. PAT (Advanced Level) Practise - 1098. Insertion or Heap Sort (25)

    http://www.patest.cn/contests/pat-a-practise/1098 According to Wikipedia: Insertion sort iterates, c ...

  7. PAT (Advanced Level) 1098. Insertion or Heap Sort (25)

    简单题.判断一下是插排还是堆排. #include<cstdio> #include<cstring> #include<cmath> #include<ve ...

  8. PAT甲题题解1098. Insertion or Heap Sort (25)-(插入排序和堆排序)

    题目就是给两个序列,第一个是排序前的,第二个是排序中的,判断它是采用插入排序还是堆排序,并且输出下一次操作后的序列. 插入排序的特点就是,前面是从小到大排列的,后面就与原序列相同. 堆排序的特点就是, ...

  9. 【PAT甲级】1098 Insertion or Heap Sort (25 分)

    题意: 输入一个正整数N(<=100),接着输入两行N个数,表示原数组和经过一定次数排序后的数组.判断是经过插入排序还是堆排序并输出再次经过该排序后的数组(数据保证答案唯一). AAAAAcce ...

随机推荐

  1. 用R语言完成的交通可视化报告

    http://sztocc.sztb.gov.cn/roadcongmore.aspx最终实现这几个图:1. 实时道路交通可视化2. 实时道路拥堵排名3. 历史路况时间序列图4. 每日每小时道况热力图 ...

  2. [Python] partial改变方法默认參数

    Python 标准库中 functools库中有非常多对方法非常有有操作的封装,partial Objects就是当中之中的一个,他是对方法參数默认值的改动. 以下就看下简单的应用測试. #!/usr ...

  3. Android 信息提示——Toast方式

    Toast用于向用户显示一些帮助/提示.一下列举了5中样式. 一.默认的效果(显示在屏幕的底部) 代码: Toast.makeText(getApplicationContext(), "默 ...

  4. 51nod 1353 树

    树背包 设f[i][j]表示第i个点,和子节点组成的联通块大小为j,其他都可行的方案 j=0表示可行的总方案 #include<cstdio> #include<iostream&g ...

  5. ssh连接超时问题解决方案,每一种方案都可以

    1.服务端修改 vim /etc/ssh/sshd_config 修改 ClientAliveInterval 60 ClientAliveCountMax 40 60秒,向客户端发送一次请求. 超过 ...

  6. poj 2104 K-th Number 主席树+超级详细解释

    poj 2104 K-th Number 主席树+超级详细解释 传送门:K-th Number 题目大意:给出一段数列,让你求[L,R]区间内第几大的数字! 在这里先介绍一下主席树! 如果想了解什么是 ...

  7. Bing Maps进阶系列八:在Bing Maps中集成OpenStreetMap地图

    Bing Maps进阶系列八:在Bing Maps中集成OpenStreetMap地图 OSM(OpenStreetMap-开放街道地图)服务就是一种发布自己地图数据图片为服务的一种实现类型,开放街道 ...

  8. CodeForces - 557D Vitaly and Cycle(二分图)

    Vitaly and Cycle time limit per test 1 second memory limit per test 256 megabytes input standard inp ...

  9. JVM面试总结

    1.  Java虚拟机的内存布局(运行时数据区) 参考:https://www.cnblogs.com/lostyears/articles/8984171.html 2. GC算法及几种垃圾收集器 ...

  10. NetCore Netty 框架 BT.Netty.RPC 系列随讲 二 WHO AM I 之 NETTY/NETTY 与 网络通讯 IO 模型之关系?

    一:NETTY 是什么? Netty 是什么?  这个问题其实百度上一搜一堆. 这是官方话的描述:Netty 是一个基于NIO的客户.服务器端编程框架,使用Netty 可以确保你快速和简单的开发出一个 ...