A1089. Insert or Merge (25)

1089. Insert or Merge (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

103 1 2 8 7 5 9 4 6 01 2 3 7 8 5 9 4 6 0

Sample Output 1:

Insertion Sort1 2 3 5 7 8 9 4 6 0

Sample Input 2:

103 1 2 8 7 5 9 4 0 61 3 2 8 5 7 4 9 0 6

Sample Output 2:

Merge Sort1 2 3 8 4 5 7 9 0 6

此题要注意中间序列与初始序列相同情况

可利用插入排序特征：即未排序序列与原序列未排序序列相同

下面归并排序想直接用讨巧方法做，找中间序列未排序序列首元素以此为步长，进行一次归并排序

但是最后一个测试点总是过不去 ， 然后直接从原始序列一步归并一步并比较的做，最后一个测试点能过

#include<cstdio>#include<algorithm>using namespace std;const int maxn = 105;int a[maxn], b[maxn];int main(){  int i, n;  scanf("%d", &n);  for(i = 0; i < n; ++i)    scanf("%d", &a[i]);  for(i = 0; i < n; ++i)    scanf("%d", &b[i]);  int start = 1;  while(start < n && b[start - 1] <= b[start]) ++start;  int p = start;  while(p < n && a[p] == b[p]) ++p;//利用插入排序特征：即未排序序列与原序列相同  if(p == n){    printf("Insertion Sort\n");    sort(a, a + start + 1);  }else{    printf("Merge Sort\n");    int step = 1, flag = 1;  //下面归并排序直接用讨巧方法做，找b数组未排序序列首元素以此为步长，进行一次归并排序//但是最后一个测试点总是过不去 ， 然后直接从原始数组一步归并一步并比较的做，最后一个测试点能过/*int step = 2 * start;for(int j = 0; j  < n; j += step){  sort(b + j, b + min(j + step, n));}*/    while(flag && step <= n)     {                              flag = 0;      for(i = 0; i < n; ++i)        if(a[i] != b[i])          flag = 1;      for(i = 0; i < n; i += step)        sort(a + i, a + min(i + step, n));      step *= 2;    }  }  for(i = 0; i < n; ++i)    printf("%s%d", i ? " " : "", a[i]);  return 0;}