题目链接：http://www.patest.cn/contests/pat-a-practise/1089
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 resulting 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：
给出两个序列，第一个是初始序列，第二个是用某种排序来排序的过程中的一个序列！
让我们判断第二个序列是用的哪一种方法排序的，并且输出第二个序列用这种排序来排序的下一个序列！
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int a[147], b[147];
int n;
int isequal(int x)
{
for(int i = x+1; i <= n; i++)
{
if(a[i] != b[i])
{
return 0;
}
}
return 1;
}
int SORT(int len)
{
for(int i = 1; i <= n; i+=len)
{
for(int j = i; j < n && j < i+len-1; j++)
{
if(b[j] > b[j+1])
{
return 0;
}
}
}
return 1;
}
int main()
{
while(~scanf("%d",&n))
{
int i;
for(i = 1; i <= n; i++)
{
scanf("%d",&a[i]);
}
for(i = 1; i <= n; i++)
{
scanf("%d",&b[i]);
}
for(i = 1; i < n; i++)
{
if(b[i] > b[i+1])
{
break;
}
}
if(isequal(i+1))//后面的都相等
{
printf("Insertion Sort\n");
sort(b+1,b+i+1 +1);
printf("%d",b[1]);
for(i = 2; i <= n; i++)
{
printf(" %d",b[i]);
}
printf("\n");
}
else
{
printf("Merge Sort\n");
int len = 2;
while(SORT(len))//寻找归并排序当前的子区间长度
{
len *=2;
}
for(i = 1; i <= n; i+=len)
{
if(i+len <= n)
{
sort(b+i,b+i+len);
}
else
{
sort(b+i,b+n+1);
}
}
printf("%d",b[1]);
for(i = 2; i <= n; i++)
{
printf(" %d",b[i]);
}
printf("\n");
}
}
return 0;
}
/*
10
3 1 2 8 7 5 9 4 6 0
1 2 3 7 8 5 9 4 6 0
10
3 1 2 8 7 5 9 4 0 6
1 3 2 8 5 7 4 9 0 6
10
3 1 2 8 7 5 9 4 0 6
1 3 2 8 7 5 9 4 0 6
*/