AtCoder

D - Median of Medians

Time limit : 2sec / Memory limit : 1024MB

Score : 700 points
Problem Statement

We will define the median of a sequence b of length M, as follows:

    Let b' be the sequence obtained by sorting b in non-decreasing order. Then, the value of the (M⁄2+1)-th element of b' is the median of b. Here, ⁄ is integer division, rounding down.

For example, the median of (10,30,20) is 20; the median of (10,30,20,40) is 30; the median of (10,10,10,20,30) is 10.

Snuke comes up with the following problem.

You are given a sequence a of length N. For each pair (l,r) (1≤l≤r≤N), let ml,r be the median of the contiguous subsequence (al,al+1,…,ar) of a. We will list ml,r for all pairs (l,r) to create a new sequence m. Find the median of m.
Constraints

    1≤N≤105
    ai is an integer.
    1≤ai≤109

Input

Input is given from Standard Input in the following format:

N
a1 a2 … aN

Output

Print the median of m.
Sample Input 1
Copy

3
10 30 20

Sample Output 1
Copy

30

The median of each contiguous subsequence of a is as follows:

    The median of (10) is 10.
    The median of (30) is 30.
    The median of (20) is 20.
    The median of (10,30) is 30.
    The median of (30,20) is 30.
    The median of (10,30,20) is 20.

Thus, m=(10,30,20,30,30,20) and the median of m is 30.
Sample Input 2
Copy

1
10

Sample Output 2
Copy

10

Sample Input 3
Copy

10
5 9 5 9 8 9 3 5 4 3

Sample Output 3
Copy

8

题意：给出一个序列，求每个子区间中中位数的中位数

题解：

首先如果A过牛客第一场提高组集训T1的神仙们肯定会知道前缀和+二分搞所有区间中位数的玩法

大致操作如下：

如果一个数比mid大，把它赋值为-1否则为1

对这个数组来个前缀和

此时如果某个区间的和为正数说明这个区间的中位数比mid小

然后问题转化成如何求现在有多少个这个区间

显然对前缀和求个顺序对就可以啦

接着就是二分答案

复杂度nlognlogai

代码如下：

#include<bits/stdc++.h>
#define lson root<<1
#define rson root<<1|1
using namespace std;
 
int n;
int a[200010];
int sum[200010];
 
struct node
{
    int l,r,sum;
}tr[1600080];
 
int push_up(int root)
{
    tr[root].sum=tr[lson].sum+tr[rson].sum;
}
 
int build(int root,int l,int r)
{
    if(l==r)
    {
        tr[root].sum=0;
        tr[root].l=l;
        tr[root].r=r;
        return 0;
    }
    tr[root].l=l;
    tr[root].r=r;
    int mid=(l+r)>>1;
    build(lson,l,mid);
    build(rson,mid+1,r);
    push_up(root);
}
 
int update(int root,int pos)
{
    if(pos==tr[root].l&&pos==tr[root].r)
    {
        tr[root].sum++;
        return 0;
    }
    int mid=(tr[root].l+tr[root].r)>>1;
    if(pos<=mid)
    {
        update(lson,pos);
    }
    else
    {
        update(rson,pos);
    }
    push_up(root);
}
 
int query(int root,int l,int r)
{
    if(l>r) return 0;
    if(l<=tr[root].l&&tr[root].r<=r) return tr[root].sum;
    int mid=(tr[root].l+tr[root].r)>>1;
    if(r<=mid)
    {
        return query(lson,l,r);
    }
    else
    {
        if(l>mid)
        {
            return query(rson,l,r);
        }
        else
        {
            return query(lson,l,mid)+query(rson,mid+1,r);
        }
    }
}
 
int check(int x)
{
    long long ans=0;
    memset(sum,0,sizeof(sum));
    for(int i=1;i<=n;i++)
    {
        sum[i]=sum[i-1]+((a[i]<=x)?1:-1);
    }
    for(int i=0;i<=n;i++)
    {
        sum[i]+=n+1;
    }
    build(1,1,200010);
    for(int i=0;i<=n;i++)
    {
        ans+=query(1,1,sum[i]-1);
        update(1,sum[i]);
    }
    return ans>=1ll*n*(n+1)/4+1;
}
 
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    int l=1,r=1e9,mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(check(mid))
        {
            r=mid;
        }
        else
        {
            l=mid+1;
        }
        if(r-l<=1) 
        {
            mid=check(l)?l:r;
            break;
        }
    }
    printf("%d
",mid);
}