題目描述

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

考點(diǎn)

插入排序、歸并排序

思路

1.判斷是哪種排序
插入排序的前n項(xiàng)一定是由小到大排列的，剩余的項(xiàng)保持原來(lái)的位置。因此只需判斷出插入排序，否則就是堆排序。
2.再進(jìn)行一次排列
插入排序：將n+1項(xiàng)重排即可；
歸并排序：按照歸并排序的步驟進(jìn)行直到某次待排串與題目給出的串相同，再進(jìn)行一次歸并即可。

代碼

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) cin >> b[i];
    int p = 1, index = 0;
    while (p < n && b[p] >= b[p - 1]) p++;
    index = p;
    while (p < n&&a[p] == b[p]) p++;
    if (p == n) {
        cout << "Insertion Sort\n";
        sort(a.begin(), a.begin() + index + 1);
    }
    else {
        cout << "Merge Sort\n";
        int k = 1, flag = 1;
        while (flag) {
            flag = 0;
            for (int i = 0; i < n; i++)
                if (a[i] != b[i]) flag = 1;
            k *= 2;
            for (int i = 0; i < n / k; i++)
                sort(a.begin() + i*k, a.begin() + (i + 1)*k);
            sort(a.begin() + n / k*k, a.begin() + n);
        }
    }
    for (int i = 0; i < n; i++)
        cout << (i == 0 ? "" : " ") << a[i];
    return 0;
}