PTA 09-排序2 Insert or Merge (25分)

Posted 2020-09-27 gravitykey

题目地址

https://pta.patest.cn/pta/test/16/exam/4/question/675

 

5-13 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 NN (\le 100≤100). Then in the next line, NN integers are given as the initial sequence. The last line contains the partially sorted sequence of the NN 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 "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

没什么好解法，只能一边排一边试

/*
评测结果
时间	结果	得分	题目	编译器	用时（ms）	内存（MB）	用户
2017-07-06 00:40	正在评测	0	5-13	gcc	无	无	
测试点结果
测试点	结果	得分/满分	用时（ms）	内存（MB）
测试点1	答案正确	7/7	1	1
测试点2	答案正确	6/6	2	1
测试点3	答案正确	1/1	1	1
测试点4	答案正确	1/1	2	1
测试点5	答案正确	3/3	1	1
测试点6	答案正确	4/4	1	1
测试点7	答案正确	3/3	2	1

用自顶向下递归法做归并被坑了，这题需要自底向上两两合并
*/

#include<stdio.h>
#define MAXN 100
int A[MAXN],B[MAXN],tmp[MAXN];
int gNextTurnToPrint=0;
int gPrinted=0;
int gSortType=0;
void swap(int *a,int *b)
{
	int temp;
	temp=*a;
	*a=*b;
	*b=temp;
}

void CheckMethod(int N)
{
	if(gPrinted)
		return;
		
	int i;
	if (gNextTurnToPrint==1)
	{
		if(gSortType==0)
			printf("Insertion Sort\n");
		else
			printf("Merge Sort\n");
			
		for(i=0;i<N;i++)
			{
				printf("%d",A[i]);
				if(i!=N-1)
					printf(" ");
			}
		gNextTurnToPrint=0;
		gPrinted=1;
		return;
	}
	
	for (i=0;i<N;i++)
	{
		if(A[i]!=B[i])
		return;
	}
	gNextTurnToPrint=1;
}

void InsertionSort(int a[],int left ,int right,int N)
{
	int i,j,temp;
	for(i=left;i<right;i++)
	{
		temp=a[i+1];
		for(j=i+1;j>left;j--)
		{
			if(temp<a[j-1])
				a[j]=a[j-1];
			else break;
		}
		a[j]=temp;
		CheckMethod(N);
	}
}
void Merge(int a[],int tmp[],int left,int leftend,int right,int rightend)
{
	int i,j,k;
	i=left;
	j=right;
	k=left;
	while(i<=leftend && j<=rightend)
	{
		if (a[i]<=a[j])
			tmp[k++]=a[i++];
		else
			tmp[k++]=a[j++];
	}
	while(i<=leftend)
		tmp[k++]=a[i++];
	while(j<=rightend)
		tmp[k++]=a[j++];

	k=left;
	while(k<=rightend)
	{
		a[k]=tmp[k];
		k++;
	}
}

void MergeSort(int a[],int tmp[],int left,int right,int N)
{
	int mid=(left+right)/2;
	if(left<right)
	{
		MergeSort(a,tmp,left,mid,N);
		MergeSort(a,tmp,mid+1,right,N);
		Merge(a,tmp,left,mid,mid+1,right);
	}
	CheckMethod(N);
}

void Merge_pass(int a[],int tmp[],int N)
{
	int i,length=1;
	while(length<N)
	{
		for(i=0;i<N-2*length;i+=2*length)
			Merge(a,tmp,i,i+length-1,i+length,i+2*length-1);
		if(i+length<N)
			Merge(a,tmp,i,i+length-1,i+length,N-1);
		length*=2;
		CheckMethod(N);
	}
}
int main()
{
	int i,N;
	scanf("%d",&N);
	
	for(i=0;i<N;i++)
	{
		scanf("%d",&A[i]);
		tmp[i]=A[i];
	}

	for(i=0;i<N;i++)
	{
		scanf("%d",&B[i]);
	}
	
	InsertionSort(A,0,N-1,N);
	
	if(!gPrinted)
	{
		for(i=0;i<N;i++)
		{
			A[i]=tmp[i];
		}
		gSortType=1;
		Merge_pass(A,tmp,N);
	}

}