PAT 1098. Insertion or Heap Sort (25) 又掉出题人的语言陷阱里了！！

1098. Insertion or Heap Sort (25)

时间限制

100 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. At 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:

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 6 06 4 5 1 0 3 2 7 8 9

Sample Output 2:

Heap Sort5 4 3 1 0 2 6 7 8 9

允许负数！！允许重复数字！！！我向来不惮以最坏的恶意揣度出题人，然而我还不料她会凶残到如此地步。

好气啊。。。ZZZ。

然后只要会堆排序，这题应该没问题。

#include <iostream>#include<stdio.h>#include<string>#include<queue>#include<algorithm>#include<string.h>#include<vector>#include<map>#include<math.h>using namespace std;int a[105]={0},b[105]={0};int n;int pick;int p;int judge() {     p=1;     while(b[p]<=b[p+1]&&p<n) p++;     sort(a+1,a+p+1);     for(int i=1;i<=n;i++)     if(a[i]!=b[i]) return 1;     return 0; }//void check(int x)//{//    if(x>pick) return;//    int z1;//    int z2;//    if(x*2>pick) z1=z2=-1;//    else z1=b[x*2];//    if(x*2+1>pick) z2=-1;//    else z2=b[x*2+1];//    int maxx=-1;//    maxx=max(z1,z2);//    int flag=1;//    if(maxx==b[x*2]) flag=0;//    if(b[x]<maxx)//    {//        if(flag==1) {swap(b[x],b[x*2+1]);check(x*2+1);}//        else//        {//           swap(b[x],b[2*x]);check(x*2);//        } //   }//}void check(int x) {     int tmp=b[1];    int root=1;    for(int i=2;i<pick;i=2*i)    {        if(i+1<pick&&b[i]<b[i+1])            ++i;        if(tmp>=b[i])            break;        else        {            b[root]=b[i];            root=i;        }    }    b[root]=tmp;}int main(){   cin>>n;   for(int i=1;i<=n;i++)    cin>>a[i];   for(int i=1;i<=n;i++)    cin>>b[i];    int x=judge();     if(x)  //heap     {         cout<<"Heap Sort"<<endl;           pick=n;         while(b[1]<b[pick]&&pick>1) pick--;         swap(b[1],b[pick]);         check(1);     }    else    {        cout<<"Insertion Sort"<<endl;        if(p==n) p--;         sort(b+1,b+1+p+1);    }       cout<<b[1];       for(int i=2;i<=n;i++)        cout<<' '<<b[i];    return 0;}