1089 insert or merge, including test point 5

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 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 "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

Thank the user Muzi Supplementary data ！

Code length limit

16 KB

The time limit

200 ms

Memory limit

The main idea of the topic ： Determine whether to merge sort or insert sort .

Insert sorting to judge , The principle is to sort the first few elements according to this , So the elements after insertion and sorting are not arranged well , So we can judge whether it is insertion sort based on this , Either insert sort or merge sort , Merge and sort , Just follow the simulation , In fact, both can be simulated , Enough time .

Code ：

#include <bits/stdc++.h>
using namespace std;
int a[101];
int b[101];
int n;
int work[101];
bool issame(){// Determine whether the array is the same 
    int i;
    for(i=0;i<n;i++){
        if(b[i]!=work[i])return false;
    }
    return true;
}
void merge(int l,int r,int sz){// Merge sort l+sz, Is the range of the left sequence .
    int i;
    int start=l;
    int temp[n];
    for(i=0;i<n;i++)temp[i]=work[i];
    int k=min(l+sz,r-1);
    int u=min(r+sz,n-1);
    while(l<=k&&r<=u){
        if(temp[l]<=temp[r])work[start++]=temp[l++];
        else work[start++]=temp[r++];
    }
    while(l<=k){
        work[start++]=temp[l++];
    }
    while(r<=u){
        work[start++]=temp[r++];
    }
    return ;
}
int main(){
    scanf("%d",&n);
    int i;
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
        work[i]=a[i];
    }
    for(i=0;i<n;i++){
        scanf("%d",&b[i]);
    }
    for(i=0;b[i]<=b[i+1];i++);// Find the first different element 
    sort(work,work+i+1);
    if(issame()){
        printf("Insertion Sort\n");
        sort(work,work+i+2);
        for(i=0;i<n;i++){
            if(i!=0)printf(" ");
            printf("%d",work[i]);
        }
        system("pause");
        return 0;
    }
    printf("Merge Sort\n");
    int j;
    for(i=0;i<n;i++)work[i]=a[i];
    for(i=1;i<n/2;i*=2){//i For scale 
        for(j=0;j<n;j+=2*i){//j It's the left border 
            if(j+i<n)merge(j,j+i,i-1);
            else merge(j,(j+n)/2,(n-j)/2);// To prevent cross-border 
        }
        if(issame()){// Find the same as the original sequence and rearrange .
            i*=2;
            for(j=0;j<n;j+=2*i){
                merge(j,j+i,i-1);
            }
            for(j=0;j<n;j++){
                if(j!=0)printf(" ");
                printf("%d",work[j]);
            }
            system("pause");
            return 0;
        }
    }
    system("pause");
}

Test points in the process 5 error , The reason is that there is no discussion about the boundary , else merge(j,(j+n)/2,(n-j)/2);// To prevent cross-border , This statement solves this problem well .

Test point 5 Test samples for

10
3 1 2 8 7 5 9 4 6 0
1 2 3 4 5 7 8 9 0 6

Running results ：

Merge Sort
0 1 2 3 4 5 6 7 8 9

I found the problem through debugging , If the boundary is not restricted , Random values will appear during debugging , Test points after correction 5 That's how it works .