1089 Insert or Merge （25 分）

最新推荐文章于 2021-03-24 19:13:41 发布

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原文链接：http://www.cnblogs.com/zhanghaijie/p/10374580.html

本文深入探讨了插入排序和归并排序两种基本排序算法的工作原理。通过分析初始整数序列及其部分排序后的状态，能够识别出所使用的排序方法。文章提供了一个C++实现的例子，展示了如何通过代码来模拟这两种算法，并判断给定序列的排序方式。

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1089 Insert or Merge （25 分）

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.

Merge sort works as follows: Divide the unsorted list into N sublists, each containing 1 element (a list of 1 element is considered sorted). Then repeatedly merge two adjacent sublists to produce new sorted sublists until there is only 1 sublist remaining.

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

Output Specification:

For each test case, print in the first line either "Insertion Sort" or "Merge 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 0 6
1 3 2 8 5 7 4 9 0 6

Sample Output 2:

Merge Sort
1 2 3 8 4 5 7 9 0 6
思路：
　　1、使用sort函数，减少错误
　　2、模拟归并排序，主要是为了找出步长。

#include<iostream>
#include<string>
#include<vector>
#include<string>
#include<cstdio>
#include<cmath>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    scanf("%d",&n);
    int a[n+1];
    for(int i=1; i<n+1; i++)
        scanf("%d",&a[i]);
    int b[n+1];
    for(int i=1; i<n+1; i++)
        scanf("%d",&b[i]);
    int p=1;
    while(p<n&&b[p]<=b[p+1])
        p++;
    // printf("%d\n",p);
    int temp=p;
    int index=p+1;
    while(index<n+1&&a[index]==b[index])
        index++;
    if(index==n+1)
    {
        sort(b+1,b+p+2);
        printf("Insertion Sort\n");
        printf("%d",b[1]);
        for(int i=2; i<n+1; i++)
            printf(" %d",b[i]);
    }
    else
    {
        int k=2*p;
        printf("Merge Sort\n");
        bool flag=false;
        for(int step=2;step/2<=n;step*=2)
        {
            if(step!=2)
            {
                for(int i=1;i<n+1;i++)
                {
                    if(a[i]!=b[i])
                        break;
                    if(i==n)
                        flag=true;
                }
            }
            for(int i=1;i<n+1;i+=step)
            {
                sort(a+i,a+min(i+step,n+1));
            }
            if(flag)
                break;
        }


        printf("%d",a[1]);
        for(int i=2; i<n+1; i++)
            printf(" %d",a[i]);

    }
    return 0;
}

转载于:https://www.cnblogs.com/zhanghaijie/p/10374580.html