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 0Sample Output 1:
Insertion Sort 1 2 3 5 7 8 9 4 6 0Sample Input 2:
10 3 1 2 8 7 5 9 4 0 6 1 3 2 8 5 7 4 9 0 6Sample Output 2:
Merge Sort 1 2 3 8 4 5 7 9 0 6
#include<cstdio>
#include<stack>
#include<iostream>
#include<vector>
#include<cstdio>
using namespace std;
int min(int a, int b){
return a > b ? b : a;
}
bool same(const vector<int>&V1, const vector<int>&V2){//比較兩個vector中的元素是否相同
for (int i = 0; i < V1.size(); ++i){
if (V1[i] != V2[i])
return false;
}
return true;
}
vector<int> merge2Vectors(vector<int>::iterator V1_begin, vector<int>::iterator V1_end, vector<int>::iterator V2_begin, vector<int>::iterator V2_end){
//合併兩個vector爲一個，歸併排序中merge的部分，其中每個vector都要是遞增的
vector<int>::iterator itr1 = V1_begin;
vector<int>::iterator itr2 = V2_begin;
vector<int>ret;
while (itr1 != V1_end && itr2 != V2_end){
if (*itr1 < *itr2){
ret.push_back(*itr1);
itr1++;
}
else{
ret.push_back(*itr2);
itr2++;
}
}
while (itr1 != V1_end){
ret.push_back(*itr1);
itr1++;
}
while (itr2 != V2_end){
ret.push_back(*itr2);
itr2++;
}
return ret;
}
vector<vector<int>>InsertSort_V;//存放插入排序每次的序列變化
vector<vector<int>>MergeSort_V;//存放歸併排序每次的序列變化
void do_InsertSort(vector<int>original){//獲取插入排序的每一次序列
int length = original.size();
for (int i = 1; i < length; ++i){//插入排序
int j = i - 1;
while (j >= 0 && original[j] > original[i]) --j;
int temp = original[i];
for (int k = i; k > j + 1; --k){
original[k] = original[k - 1];
}
original[j + 1] = temp;
if (InsertSort_V.size() == 0 || !same(InsertSort_V[InsertSort_V.size() - 1],original))
//如果序列集爲空，或者本次算法後序列有變化，才加入序列集
InsertSort_V.push_back(original);
}
}
void do_MergeSort(vector<int>original){//獲取歸併排序的每一次序列
//這裏採用的是上述的歸併排序方法2
for (int i = 1; i < original.size(); i *= 2){//i是每次的步長，每次指數2遞增
vector<int>temp;
for (int j = 0; j < original.size(); j += 2*i){//j是每組其實排序的位置，下一組就是j + 2*i
vector<int>temp_sec = merge2Vectors(original.begin() + j, original.begin() +min(j + i, original.size()), original.begin() + min(j + i, original.size()), original.begin() + min(j + 2*i, original.size()));
//注意當到達末尾的情況，需要進行一個min的判斷
temp.insert(temp.end(), temp_sec.begin(), temp_sec.end());//把每兩個小組排好序的合併到temp中
}
if (MergeSort_V.size() == 0 || !same(MergeSort_V[MergeSort_V.size() - 1], temp))
//如果序列集爲空，或者本次算法後序列有變化，才加入序列集
MergeSort_V.push_back(temp);
original = temp;
temp.clear();
}
}
void print(const vector<int>&V){//格式化打印序列
for (int i = 0; i < V.size(); ++i){
if (i == 0)cout << V[i];
else cout << " " << V[i];
}
cout << endl;
}
int main(){
freopen("F://Temp/input.txt", "r",stdin);
int n;
cin >> n;
vector<int>input;
vector<int>result;
for (int i = 0; i < n; ++i){
int num;
cin >> num;
input.push_back(num);
}
for (int i = 0; i < n; ++i){
int num;
cin >> num;
result.push_back(num);
}
do_InsertSort(input);
for (int i = 0; i < InsertSort_V.size(); ++i){
if (same(InsertSort_V[i], result)){
cout << "Insertion Sort" << endl;
if (i == InsertSort_V.size() - 1)//如果是最後一個序列相等，那麼下一個序列還是最後一個序列
print(InsertSort_V[i]);
else
print(InsertSort_V[i + 1]);
return 0;
}
}
do_MergeSort(input);
for (int i = 0; i < MergeSort_V.size(); ++i){
if (same(MergeSort_V[i], result)){
cout << "Merge Sort" << endl;
if (i == MergeSort_V.size() - 1)//如果是最後一個序列相等，那麼下一個序列還是最後一個序列
print(MergeSort_V[i]);
else
print(MergeSort_V[i + 1]);
return 0;
}
}
return 0;
}
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 101;
int num1[N];
int num2[N];
int main(void)
{
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i)
scanf("%d", &num1[i]);
for (int i = 0; i < n; ++i)
scanf("%d", &num2[i]);
bool isInsert = true;
int i = 1;
while (i < n && num2[i] >= num2[i - 1])
++i;
int j = i;
while (i < n && num2[i] == num1[i])
++i;
if (i != n)
isInsert = false;
if (isInsert)
printf("Insertion Sort\n");
else
printf("Merge Sort\n");
if (isInsert)
{
while (j > 0 && num2[j - 1] > num2[j])
{
int temp = num2[j - 1];
num2[j - 1] = num2[j];
num2[j] = temp;
--j;
}
}
else
{
int k = n;
i = 0;
while (i < n)
{
int currLength = 1;
j = i + 1;
while (j < n && num2[j - 1] <= num2[j])
{
++j;
currLength++;
}
if (j == n)
break;
else
{
k = min(currLength, k);
i = j;
}
}
for (i = 0; i < n; i += 2 * k)
{
sort(num2 + i, num2 + min(i + 2 * k, n));
}
}
printf("%d", num2[0]);
for (int i = 1; i < n; ++i)
printf(" %d", num2[i]);
printf("\n");
return 0;
}