PAT甲级 1057 Stack(30) (树状数组+二分+栈)

题目

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.

输入

Each input file contains one test case. For each case, the first line contains a positive integer N (≤). 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 .

输出

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.

样例输入 

17
Pop
PeekMedian
Push 3
PeekMedian
Push 2
PeekMedian
Push 1
PeekMedian
Pop
Pop
Push 5
Push 4
PeekMedian
Pop
Pop
Pop
Pop

样例输出 

Invalid
Invalid
3
2
2
1
2
4
4
5
3
Invalid

题意理解

让你维护一个栈 入栈和出栈没啥好说的 还有一个新的操作

就是让你找到栈的里面所有元素的个数的中位数元素是什么

元素个数为奇数就是找到（N+1）/2 个元素是什么

偶数就是 N/2

其实由于C++的特性 小技巧就是 无论奇数偶数（N+1）/2 是一样的因为是向下取整的

所有直接用 （N+1）/2 代表中位数即可

那么我们怎么维护中位数呢

我们用一个树状数组 来记录个数 树状数组本来是维护前缀和 此时我们维护这个数的出现个数

那么我们怎么找中位数呢 我们用二分 找到出现前面出现数的次数是 （N+1）/2 

那么那个数就是中位数

check条件就是如果出现次数大于等于k那么右区间缩过来 

代码 

#include
using namespace std;
const int N=2e5+10;
const int INF=0x3f3f3f3f;
stackst;
int n,m;
int tr[N];
int lowbit(int x){
    return x&-x;
}
int sum(int x){
    int res=0;
    for(int i=x;i;i-=lowbit(i))res+=tr[i];
    return res;
}
void add(int x,int c){
    for(int i=x;i>1;
        if(sum(mid)>=k)r=mid;
        else l=mid+1;
    }
    return l;
}
           
int main(){
    scanf("%d",&m);
    char op[25];
    while(m--){
        scanf("%s",op);
        if(op[1]=='o'){
            //pop
            if(!st.size()){
                puts("Invalid");
                continue;
            }
            else {
                int x=st.top();
                st.pop();
                add(x,-1);
                printf("%d\n",x);
            }
        }
        else if(op[1]=='u'){
            //push
            int x;
            scanf("%d",&x);
            st.push(x);
            add(x,1);
        }
        else if(op[1]=='e'){
            //peekmedian
            if(!st.size()){
                puts("Invalid");
                continue;
            }
            else {
                int t=find();
                printf("%d\n",t);
            }
        }
    }
    return 0;
}