【HDOJ6621】K-th Closest Distance（主席树，二分）

题意：给定一个长为n的序列，有m次强制在线的询问，每次询问位置【L,R】中abs(a[i]-p)第k小的值

n,m<=1e5，a[i]<=1e6，p<=1e6，k<=169

思路：主席树外面套个二分

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 #define N  1100000
 #define M  4100000
 #define fi first
 #define se second
 #define MP make_pair
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const ll MOD=,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;

 struct node
 {
     int l,r,s;
 }t[N*];

 int root[N],cnt;

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void update(int l,int r,int x,int &p)
 {
     t[++cnt]=t[p];
     p=cnt;
     t[p].s++;
     if(l==r) return;
     int mid=(l+r)>>;
     if(x<=mid) update(l,mid,x,t[p].l);
      else update(mid+,r,x,t[p].r);
 }

 int query(int l,int r,int x,int y,int p1,int p2)
 {
     if(x<=l&&r<=y) return t[p2].s-t[p1].s;
     int mid=(l+r)>>;
     int tmp=;
     if(x<=mid) tmp+=query(l,mid,x,y,t[p1].l,t[p2].l);
     if(y>mid) tmp+=query(mid+,r,x,y,t[p1].r,t[p2].r);
     return tmp;
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     int cas=read();
     while(cas--)
     {
         int n=read(),q=read();
         rep(i,,cnt) t[i].l=t[i].r=t[i].s=;
         cnt=; root[]=;
         rep(i,,n)
         {
             int x=read();
             root[i]=root[i-];
             update(,1e6,x,root[i]);
         }
         int ans=;
         rep(i,,q)
         {
             int l=read(),r=read(),p=read(),k=read();
             l^=ans; r^=ans; p^=ans; k^=ans;
             int left=,right=1e6;
             ans=1e6;
             while(left<=right)
             {
                 int mid=(left+right)>>;
                 int x=max(,p-mid),y=min(,p+mid);
                 int t=query(,1e6,x,y,root[l-],root[r]);
                 if(t>=k){ans=mid; right=mid-;}
                  else left=mid+;
             }
             printf("%d\n",ans);
         }
     }
     return ;
 }