本体让求第K大的答案，于是80%为二分答案，首先个时间复杂度加上一个logn。应该只能用O(logn * n)的时间复杂。首先二分答案，求当每一个数为中位数的时候的区间总数。首先每次枚举区间中有那些数大于它就赋值为1，否则赋值为-1.用树状数组或前缀和来维护区间，求大小为零的区间。相当与求顺序对的数量，即可用log的级别来求了。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pr;
const double pi=acos(-1);
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define Rep(i,u) for(int i=head[u];i;i=Next[i])
#define clr(a) memset(a,0,sizeof a)
#define pb push_back
#define mp make_pair
#define fi first
#define sc second
ld eps=1e-9;
ll pp=1000000007;
ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;}
ll powmod(ll a,ll b,ll pp){ll ans=1;for(;b;b>>=1,a=mo(a*a,pp))if(b&1)ans=mo(ans*a,pp);return ans;}
ll read(){
ll ans=0;
char last=' ',ch=getchar();
while(ch<'0' || ch>'9')last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
if(last=='-')ans=-ans;
return ans;
}
//head
#define N 110000
int a[N],b[N],c[N],d[N],e[N],n,m;
ll k;
int lowbit(int x){
return x&(-x);
}
int bin(int k){
int l=1,r=m;
while(l<r){
int mid=(l+r)/2;
if(k<=d[mid])r=mid;
else l=mid+1;
}
return l;
}
void add(int x){
for(;x<=m;x+=lowbit(x))e[x]++;
}
int find(int x){
int ans=0;
for(;x;x-=lowbit(x))ans+=e[x];
return ans;
}
ll check(int k){
c[0]=0;
rep(i,1,n)c[i]=c[i-1]+(a[i]>=k);
rep(i,0,n)c[i]=c[i]*2-i,d[i+1]=c[i];
sort(d+1,d+n+2);
d[0]=1;
rep(i,2,n+1)
if(d[i]!=d[d[0]])d[++d[0]]=d[i];
m=d[0];
ll ans=0;
rep(i,0,n)c[i]=bin(c[i]);
rep(i,0,m)e[i]=0;
rep(i,0,n)
if(i&1)add(c[i]);
else ans+=find(c[i]-1);
rep(i,0,m)e[i]=0;
rep(i,0,n)
if((i&1)==0)add(c[i]);
else ans+=find(c[i]-1);
return ans;
}
int main(){
n=read();k=read();
rep(i,1,n)a[i]=read(),b[i]=a[i];
sort(b+1,b+n+1);
b[0]=1;
rep(i,2,n)
if(b[i]!=b[b[0]])b[++b[0]]=b[i];
int l=1,r=b[0];
while(l<r){
int mid=(l+r)/2+1;
ll tt=check(b[mid]);
if(tt==k){
cout<<b[mid]<<endl;
return 0;
}
if(check(b[mid])>k)l=mid;
else r=mid-1;
}
cout<<b[l]<<endl;
return 0;
}