Pat class a a1057 stack

PAT Class A A1057

1057 Stack (30 branch )
Stack is one of the most fundamental data structures, which is based on the principle of Last In First Out (LIFO). The basic operations include Push (inserting an element onto the top position) and Pop (deleting the top element). Now you are supposed to implement a stack with an extra operation: PeekMedian – return the median value of all the elements in the stack. With N elements, the median value is defined to be the (N/2)-th smallest element if N is even, or ((N+1)/2)-th if N is odd.

Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤10
​5
​​ ). Then N lines follow, each contains a command in one of the following 3 formats:

Push key
Pop
PeekMedian
where key is a positive integer no more than 10
​5
​​ .

Output Specification:
For each Push command, insert key into the stack and output nothing. For each Pop or PeekMedian command, print in a line the corresponding returned value. If the command is invalid, print Invalid instead.

Sample Input:
17
Pop
PeekMedian
Push 3
PeekMedian
Push 2
PeekMedian
Push 1
PeekMedian
Pop
Pop
Push 5
Push 4
PeekMedian
Pop
Pop
Pop
Pop
Sample Output:
Invalid
Invalid
3
2
2
1
2
4
4
5
3
Invalid

    thought ： For sequences that update data in real time , The... In a single query sequence K A large number will cause O(n) Time complexity of ,
    But you add sorting , This is again. O(n*lgn), Of course , Do not reorder , Update the original sequence directly , A total of
    O(n) Time complexity of ;
    Block thought ： Block the number of queries , Each block stores a certain range of values respectively , Then divided into ceil(n^0.5) block , Each piece
    It is divided into floor(n^0.5), perhaps ceil(n^0.5) Number of ranges

hash Array table[i] Represents the current integer i The number of existence ,block[i] It means the first one i The number of elements in a block ;

Algorithm notes ：

/**
     thought ： For sequences that update data in real time , The... In a single query sequence K A large number will cause O(n) Time complexity of ,
     But you add sorting , This is again. O(n*lgn), Of course , Do not reorder , Update the original sequence directly , A total of 
    O(n) Time complexity of ;
     Block thought ： Block the number of queries , Each block stores a certain range of values respectively , Then divided into ceil(n^0.5) block , Each piece 
     It is divided into floor(n^0.5), perhaps ceil(n^0.5) Number of ranges 
*/

#include <iostream>
#include <cmath>
#include <stack>
#include <string>
#include <cstring>
using namespace std;
const int maxn=100001;
const int sqr_n=316;
int every_block_num(int n);// Find the maximum number of each block 
void Push(int n);
void Pop();
void PeekMedian();
stack<int> sta;
int table[maxn]; /** table[i] Represents the current integer i The number of existence */
int block[sqr_n+1]; /** block[i] It means the first one i The number of elements in a block */
int main()
{
//    cout << 316*316 << endl;
//    cout << (99999/316) <<endl;
//    while(1);
    memset(table,0,sizeof(table));
    memset(block,0,sizeof(block));
    int query_num,x;
    cin >> query_num;
    string str;
//    cout << ceil(sqrt(n*1.0)) << endl;    // Calculate the total number of blocks , How many elements are there in each block 
//    cout << every_block_num(n) << endl;
    for(int i=0;i<query_num;++i)
    {
        cin >> str;
        if(str=="Push")
        {
            cin >> x;
            Push(x);
        }
        else if(str=="Pop")
        {
            if(sta.empty())
                cout << "Invalid\n";
            else
            {
                Pop();
            }
        }
        else
        {
            if(sta.empty())
                cout << "Invalid\n";
            else
            {
                PeekMedian();
            }
        }
    }
    return 0;
}

int every_block_num(int n)
{
    double sqr=sqrt(n*1.0);
    double floor_sqr=floor(sqr);
    double ceil_sqr=ceil(sqr);
    if(floor_sqr*ceil_sqr<n)
        return ceil_sqr;
    else
        return floor_sqr;
}

void Push(int n)
{
    sta.push(n);
    ++table[n];
    ++block[n/sqr_n];
}

void Pop()
{
    int top=sta.top();
    sta.pop();
    --table[top];
    --block[top/sqr_n];
    cout << top << endl;
}

//void PeekMedian()
//{
//    int size=sta.size();
//    int k=(size+1)/2;
//    int sum=0,i;
//    for(i=0;i<316;++i)
//    {
//        if(sum+block[i]>=k)
//            break;
//        sum+=block[i];
//    }
//    int en=(i+1)*316,j;
//    for(j=i*316;j<en;++j)
//    {
//        if(sum+table[j]>=k)
//            break;
//        sum+=table[j];
//    }
//    cout << j << endl;
//}

//PeekMedian Another way of writing , use while Control cycle 
void PeekMedian()
{
    int size=sta.size();
    int k=(size+1)/2;
    int sum=0,index=0;
    while(sum+block[index]<k)
        sum+=block[index++];
    int j=index*sqr_n;
    while(sum+table[j]<k)
        sum+=table[j++];
    cout << j << endl;
}