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题意：现给定原始序列和由某排序算法产生的中间序列，请你判断该算法是插入算法还是归并算法。首先在第1行中输出“Insertion Sort”表示插入排序、或“Merge Sort”表示归并排序；然后在第2行中输出用该排序算法再迭代一轮的结果序列
思路：考察插入排序和归并排序。
根据插入排序的特点，中间序列的前面部分是非递减的，后面部分是相同的 以此来判断插入排序。
归并排序可以使用 循环+sort 模拟。
先将i指向中间序列中满足从左到右是从小到大顺序的最后一个下标，再将j指向从i+1开始，第一个不满足a[j] == b[j]的下标，如果j顺利到达了下标n，说明是插入排序，再下一次的序列是sort(a, a+i+2);否则说明是归并排序。归并排序就别考虑中间序列了，直接对原来的序列进行模拟归并时候的归并过程，i从0到n/k，每次一段段得sort(a + i * k, a + (i + 1) * k);最后别忘记还有最后剩余部分的sort(a + n / k * k, a + n);这样是一次归并的过程。直到有一次发现a的顺序和b的顺序相同，则再归并一次，然后退出循环～
以上参考柳婼 の blog
代码：
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define mem(a,n) memset(a,n,sizeof(a))
#define lowbit(i) ((i)&(-i))
typedef long long ll;
typedef unsigned long long ull;
const ll INF=0x3f3f3f3f;
const double eps = 1e-6;
const int N = 1e3+5;
//int a[N],b[N];
int main() {
int n;
cin>>n;
int *a=new int[n];
int *b=new int[n];
for(int i=0; i<n; i++) {
cin>>a[i];
}
for(int i=0; i<n; i++) {
cin>>b[i];
}
int i,j;
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;
// cout<<"k="<<k<<endl;
for(i=0; i<n/k; i++) {///sort来模拟归并排序
sort(a+i*k,a+(i+1)*k);
}
sort(a+n/k*k,a+n);///n%k!=0时
// for(int h=0;h<n;h++){///某次排序后的中间序列
// cout<<a[h]<<" ";
// }
// cout<<endl;
}
}
for(j=0; j<n; j++) {
if(j) cout<<" ";
cout<<a[j];
}
delete []a;
delete []b;
return 0;
}
