PAT1 1057 Stack

Subject to the effect

A stack is now given, then given three operations

The number x onto the stack
The top of the stack out of the stack
Asked 栈的 median

Entry

Each test case comprising a
first row is a positive integer $N\leq10^5$ , then there $N$ line command format is as follows

Push key
Pop
PeekMedian

Push keyThe keystack, Popthe top element from the stack, PeekMedianfind the median of all elements in the stack. keyIs a positive integer not more than $10^5$

Export

Each Popoutput an instruction stack elements, each PeekMedianmedian of all elements in the stack output command. If an element can not stack or the stack output is outputInvalid

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

Resolve

This problem can be used Fenwick tree to solve.
But the tree is stored in the array subscript i is the number of how many before i , and so seek prefix and then compare the size of the stack can be found with a median stack Fenwick tree
but this problem will be with python timeout, c ++ can be AC
timeout python:

# -*- coding: utf-8 -*- 
# @Time : 2019/6/13 11:10 
# @Author : ValarMorghulis 
# @File : 1057.py
s = list()
c = [0 for i in range(100861)]


def lowBit(x):
    return x & (-x)


def update(x, v):
    i = x
    while i < 100861:
        c[i] += v
        i += lowBit(i)


def getSum(x):
    sum, i = 0, x
    while i >= 1:
        sum += c[i]
        i -= lowBit(i)
    return sum


def find():
    left, right, k = 1, 100861, (len(s) + 1) // 2
    while left < right:
        mid = (left + right) // 2
        if getSum(mid) >= k:
            right = mid
        else:
            left = mid + 1
    print(left)


def solve():
    global c, s
    n = int(input())
    for i in range(n):
        command = input().split()
        if command[0][1] == 'u':
            s.append(int(command[1]))
            update(int(command[1]), 1)
        elif command[0][1] == 'o':
            if len(s) == 0:
                print("Invalid")
            else:
                update(s[-1], -1)
                print(s[-1])
                s.pop()
        elif command[0][1] == 'e':
            if len(s) == 0:
                print("Invalid")
            else:
                find()


if __name__ == "__main__":
    solve()

AC 的 c ++

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<string>
#include<cmath>

#define inf 0xffffffff
#define lowbit(i) ((i)&(-(i)))
using namespace std;


const int maxn=100861;

stack<int> s;
int c[maxn];

void update(int x, int v)
{
    for(int i=x; i<maxn; i+=lowbit(i))
        c[i]+=v;
}

int sum(int x)
{
    int t=0;
    for(int i=x; i>=1; i-=lowbit(i))
        t+=c[i];
    return t;
}

void find()
{
    int left=0;
    int right=maxn;
    int mid;
    int k=(s.size()+1)>>1;
    while(left<right)
    {
        mid=(left+right)>>1;
        if(sum(mid)>=k)
            right=mid;
        else
            left=mid+1;
    }
    printf("%d\n", left);
}

int main()
{
    int n;
    scanf("%d", &n);
    char command[15];
    for(int i=0; i<n; i++)
    {
        scanf("%s", command);
        if(command[1]=='u')
        {
            int t;
            scanf("%d", &t);
            update(t, 1);
            s.push(t);
        }
        else
            if(command[1]=='o')
                if(s.size()==0)
                    printf("Invalid\n");
                else
                {
                    update(s.top(), -1);
                    printf("%d\n", s.top());
                    s.pop();
                }
            else
                if(command[1]=='e')
                    if(s.size()==0)
                        printf("Invalid\n");
                    else
                        find();
    }
    return 0;
}