参考自柳诺https://www.liuchuo.net/

#include <iostream>
#include <algorithm>

using namespace std;
int main(){
    //输入
    int a[110],b[100];
    int n;
    cin >> n;
    for(int i = 0;i < n;i++){
        cin >> a[i];
    }
    for(int i = 0;i < n;i++){
        cin >> b[i];
    }
    //判断是否是插入算法，归并不好直接判断
    int pos = 0;
    for(int i = 0;i < n - 1;i++){ //找到
        if(b[i] > b[i+1]){
            pos = i;
            break;
        }
    }
    int j = pos + 1;
    while(j < n && a[j] == b[j])j++;
    if(j == n){
        cout << "Insertion Sort" << endl;
        sort(a,a + pos + 2);    //下一轮应排序到[0,pos + 1].一共(pos + 2)个元素
    }else{
        cout << "Merge Sort" << endl;
        int flag = 1,k = 1;
        while(flag){
            flag = 0;
            for(int i = 0;i < n;i++){ //对a[]进行归并排序，直到和b[]相等为止
                if(a[i] != b[i])
                    flag = 1;
            }    
            k = k * 2;              //归并的步长，每次x2
            for(int i = 0;i < n / k;i++){  //k被n整除部分
                sort(a + i * k,a + (i + 1) * k);
            }
            sort(a + n/k * k,a + n);  //余下部分
        }
        for(int i = 0;i < n;i++){ //上边while()里边，满足相等条件后又排序了一次，所以直接输出即可
            if(i != 0)cout << " ";
            cout << a[i];
        }
    }
    return 0;
}

柳诺的代码

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int n, a[100], b[100], i, j;
    cin >> n;
    for (int i = 0; i < n; i++)
        cin >> a[i];
    for (int i = 0; i < n; i++)
        cin >> b[i];
    for (i = 0; i < n - 1 && b[i] <= b[i + 1]; i++)
        ;
    for (j = i + 1; a[j] == b[j] && j < n; j++)
        ;
    if (j == n)
    {
        cout << "Insertion Sort" << endl;
        sort(a, a + i + 2);
    }
    else {
        cout << "Merge Sort" << endl;
        int k = 1, flag = 1;
        while (flag)
        {
            flag = 0;
            for (i = 0; i < n; i++)
            {
                if (a[i] != b[i])
                    flag = 1;
            }
            k = k * 2;
            for (i = 0; i < n / k; i++)
                sort(a + i * k, a + (i + 1) * k);
            sort(a + n / k * k, a + n);
        }
    }
    for (j = 0; j < n; j++)
    {
        if (j != 0)
            printf(" ");
        printf("%d", a[j]);
    }
    return 0;
}