题目

According to Wikipedia:

Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. 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

题目分析

已知排序前和排序一部分数据后的序列,判断是插入排序还是堆排序,并打印下一次排序的结果(已知所有序列最终排序结果为升序,隐含条件:堆为大顶堆)

解题思路

  1. 判断是插入排序还是堆排序,插入排序中间序列特点:从0i是已经排序好的,in-1跟排序前一模一样
  2. 排序

    2.1 插入排序下一次排序结果,不需要模拟插入排序,只要用sort将0~i+1升序排序即可

    2.2 堆排序下一次排序结果,需要从n-1倒序寻找第一个小于0处元素的位置k,交换0与k处元素,对堆顶元素从0~k-1向下堆化

Code

#include <iostream>
#include <algorithm>
using namespace std;
int bs[100],as[100];
void downAdjust(int low,int high) {
int i=low,j=2*i+1;
while(j<=high) {
if(j+1<=high&&as[j+1]>as[j])j=j+1;
if(as[i]<as[j]) {
swap(as[i],as[j]);
i=j;
j=2*i+1;
} else {
break;
}
}
}
int main(int argc,char * argv[]) {
int n;
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%d",&bs[i]);
for(int i=0; i<n; i++)
scanf("%d",&as[i]);
int i,j;
for(i=0; i<n&&as[i]<=as[i+1]; i++);
for(j=i+1; j<n&&as[j]==bs[j]; j++);
if(j==n) {
printf("Insertion Sort\n");
sort(as,as+i+1+1);
} else {
printf("Heap Sort\n");
for(j=n-1; j>=1&&as[j]>=as[0]; j--);
swap(as[0],as[j]);
downAdjust(0,j-1);
}
printf("%d",as[0]);
for(int i=1; i<n; i++)
printf(" %d",as[i]);
return 0;
}

PAT Advanced 1098 Insertion or Heap Sort (25) [heap sort(堆排序)]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 (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 ...

  3. 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 ...

  4. PAT (Advanced Level) 1089. Insert or Merge (25)

    简单题.模拟一下即可. #include<cstdio> #include<cstring> #include<cmath> #include<vector& ...

  5. PAT (Advanced Level) 1113. Integer Set Partition (25)

    简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...

  6. PAT (Advanced Level) 1062. Talent and Virtue (25)

    简单排序.题意较长. #include<cstdio> #include<cstring> #include<cmath> #include<queue> ...

  7. PAT (Advanced Level) 1052. Linked List Sorting (25)

    简单题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> ...

  8. PAT (Advanced Level) 1017. Queueing at Bank (25)

    简单模拟. #include<iostream> #include<cstring> #include<cmath> #include<algorithm&g ...

  9. PAT (Advanced Level) 1012. The Best Rank (25)

    简单排序题. 注意:分数相同的人排名相同. #include<iostream> #include<cstring> #include<cmath> #includ ...

随机推荐

  1. java程序题目解析

    (选择一项) A: 不能有括号 B: C: 确定最后一位 D: 正确答案是 B  本题考查的是Java数组概念,数组下标是从零开始的,但是数据下标的总量和数据长度相同 (选择一项) A: B: 顺序不 ...

  2. JMeter-响应断言设置

    针对如上请求,可以设置3种相应断言: 1. 2. 3.

  3. RDD 可视化 —— RDDOperationScope.withScope

    最近在看各种博客,学习 spark 源代码. 网上对源代码的分析基本都是基于 0.7, 0.8, 1.0 的代码,而现在的发行版已经是 1.5 了.所以有些代码不大对的上.比如函数 RDD.map() ...

  4. javascript判断数组是否包含了指定的元素

    jQuery写法: var arr = [ "xml", "html", "css", "js" ]; $.inArra ...

  5. Netty简单认识

    简介 Netty 是由JBOSS提供的一个 Java开源框架, 现在是 Github上的开源项目 Netty 是一个异步的.基于事件驱动的网络应用框架式, 用以快速开发高性能.高可靠性的网路IO程序 ...

  6. python中groupby函数详解(非常容易懂)

    一.groupby 能做什么? python中groupby函数主要的作用是进行数据的分组以及分组后地组内运算! 对于数据的分组和分组运算主要是指groupby函数的应用,具体函数的规则如下: df[ ...

  7. 《YouTube 网站的架构演进》阅读笔记

    概述 YouTube 在国内是个404网站,需要翻墙得见,这是有用的废话,先铺垫一下. 从全球网站来看,它仅次于母公司 Google,全球排名位列第2.每天超过5亿以上视频播放量,平均每个用户点击10 ...

  8. maven项目打包和运行

    1. 添加pom <build> <plugins> <plugin> <artifactId>maven-assembly-plugin</ar ...

  9. 循环指令 LOOP

    循环程序: 如果需要重复执行若干次同样任务.用循环执行 循环指令: LOOP <跳转标号> 用累加器的低字做循环计数器 每次执行LOOP 指令的时候,累加器的低字减去1 若减去后 非零 , ...

  10. POJ1088:滑雪

    滑雪 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 82112   Accepted: 30706 Description ...