题目链接
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.
Heap sort divides its input into a sorted and an unsorted region, and it iteratively shrinks the unsorted region by extracting the largest element and moving that to the sorted region. it involves the use of a heap data structure rather than a linear-time search to find the maximum.
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 "Heap 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 6 0
6 4 5 1 0 3 2 7 8 9
Sample Output 2:
Heap Sort
5 4 3 1 0 2 6 7 8 9题意：给出n和n个数的序列a和b，a为原始序列，b为排序其中的一个步骤，问b是a经过了堆排序还是插入排序的，并且输出它的下一步
类似排序题：【PAT甲级】1089 Insert or Merge（25 分）（插入/归并排序）
思路：堆排序的特点是后面是从小到大的，前面的顺序不一定，那么就可以到第一个不是前面所有序列最大值的那个数字i，把它和第一个数字交换，然后把1~i-1进行一次向下调整~~~向下调整，low和high是需要调整的区间，因为是大顶堆，就是不断比较当前结点和自己的孩子结点哪个大，如果孩子大就把孩子结点和自己交换，然后再不断调整直到到达区间的最大值不能再继续了为止
代码：
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define drep(i,n,a) for(int i=n;i>=a;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 = 1e5+5;
vector<int>a,b;
void adjust(int lo,int hi) {
int i=1,j=i*2;
while(j<=hi) {
if(j+1<=hi&&b[j]<b[j+1]) {
j++;
}
if(b[i]<b[j]) {
swap(b[i],b[j]);
i=j,j=i*2;
} else {
break;
}
}
}
int main() {
int n;
scanf("%d",&n);
a.resize(n+1),b.resize(n+1);
for(int i=1; i<=n; i++) {
scanf("%d",&a[i]);
}
for(int i=1; i<=n; i++) {
scanf("%d",&b[i]);
}
int i,j;
for(i=2; i<=n&&b[i-1]<=b[i]; i++);
for(j=i; j<=n&&a[j]==b[j]; j++);
if(j==n+1) {
printf("Insertion Sort\n");
sort(b.begin()+1,b.begin()+i+1);
} else {
printf("Heap Sort\n");
for(i=n; i>=2&&b[i]>=b[i-1]; i--);
swap(b[1],b[i]);
adjust(1,i-1);
}
for(int i=1; i<=n; i++) {
if(i!=1) printf(" ");
printf("%d",b[i]);
}
return 0;
}
