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 Nelements, 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

 題目大意：現請你實現一種特殊的堆疊，它多了一種操作叫“查中值”，即返回堆疊中所有元素的中值。對於N個元素，若N是偶數，則中值定義為第N/2個最小元；若N是奇數，則中值定義為第(N+1)/2個最小元。

純模擬一下，超時又爆棧，我就不想多說什麼了

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;
 
int n,d;;
int A[10005],B[10005];
 
int select(int num)
{
	for(int i=1;i<=d;i++) B[i]=A[i];
	sort(B+1,B+d+1);
	return B[num];
}
 
void process()
{
	char operate[1000];
	scanf("%s",operate);
	string s=operate;
 
	if(s=="Pop")
	{
		if(d==0)
			printf("Invalid\n");
		else 
		{
			printf("%d\n",A[d]);
			d--;
		}
	}
	else if(s=="PeekMedian")
	{
		if(d==0)
			printf("Invalid\n");
		else
		{
			int num;
			if(d%2==0) num=d/2;
			else
				num=(d+1)/2;
			printf("%d\n",select(num));
		}
	}
	else if(s=="Push")
	{
		int b;
		scanf("%d",&b);
		A[++d]=b;
	}
}
 
int main()
{
	//freopen("1.txt","r",stdin);
	while(scanf("%d",&n)!=EOF)
	{
		d=0;
		for(int i=0;i<n;i++) 
			process();
	}
	return 0;
}

樹狀陣列寫法

#include <iostream>
#include <stack>
#define lowbit(i) ((i) & (-i))
const int maxn = 100010;
using namespace std;
int c[maxn];
//這個是用來計數每個元素出現的次數，比如push 3,那麼就是3多了一個。
//表示≤當前下標的有幾個。
stack<int> s;
void update(int x, int v) 
{
	while(x<maxn)
	{
		c[x]+=v;
		x+=lowbit(x);
	}
}
int getsum(int x) 
{
    int sum = 0;    
    while(x>0)
	{
		sum+=c[x];
		x-=lowbit(x);
	}
    return sum;
}
void PeekMedian() {
    int left = 1, right = maxn, mid, k = (s.size() + 1) / 2;
    while(left < right) {//使用二分法查詢。
        mid = (left + right) / 2;
        if(getsum(mid) >= k)
            right = mid;
        else
            left = mid + 1;
    }
    printf("%d\n", left);
}
int main() {
    int n, temp;
    scanf("%d", &n);
    char str[15];
    for(int i = 0; i < n; i++) {
        scanf("%s", str);
        if(str[1] == 'u') {
            scanf("%d", &temp);
            s.push(temp);
            update(temp, 1);
        } else if(str[1] == 'o') {
            if(!s.empty()) {
                update(s.top(), -1);
                printf("%d\n", s.top());
                s.pop();
            } else {
                printf("Invalid\n");
            }
        } else {
            if(!s.empty())
                PeekMedian();
            else
                printf("Invalid\n");
        }
    }
    return 0;
}