hdu6621 二分加主席树

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6621

Problem Description

You have an array: a1, a2, , an and you must answer for some queries.
For each query, you are given an interval [L, R] and two numbers p and K. Your goal is to find the Kth closest distance between p and aL, aL+1, ..., aR.
The distance between p and ai is equal to |p - ai|.
For example:
A = {31, 2, 5, 45, 4 } and L = 2, R = 5, p = 3, K = 2.
|p - a2| = 1, |p - a3| = 2, |p - a4| = 42, |p - a5| = 1.
Sorted distance is {1, 1, 2, 42}. Thus, the 2nd closest distance is 1.

 

Input

The first line of the input contains an integer T (1 <= T <= 3) denoting the number of test cases.
For each test case:
冘The first line contains two integers n and m (1 <= n, m <= 10^5) denoting the size of array and number of queries.
The second line contains n space-separated integers a1, a2, ..., an (1 <= ai <= 10^6). Each value of array is unique.
Each of the next m lines contains four integers L', R', p' and K'.
From these 4 numbers, you must get a real query L, R, p, K like this:
L = L' xor X, R = R' xor X, p = p' xor X, K = K' xor X, where X is just previous answer and at the beginning, X = 0.
(1 <= L < R <= n, 1 <= p <= 10^6, 1 <= K <= 169, R - L + 1 >= K).

 

Output

For each query print a single line containing the Kth closest distance between p and aL, aL+1, ..., aR.

 

Sample Input

1

5 2

31 2 5 45 4

1 5 5 1

2 5 3 2

 

Sample Output

0

1

 

求区间内| a[i]-p | 第k小的数，二分答案，查询区间[ p-ans,p+ans ]内数的数量是否大于等于k

#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define Max 1000000
#define maxn 100005
int a[maxn],b[maxn],T[maxn<<5],L[maxn<<5],R[maxn<<5],sum[maxn<<5],tot;
inline int update(int pre,int l,int r,int x)
{
    int rt=++tot;
    L[rt]=L[pre];
    R[rt]=R[pre];
    sum[rt]=sum[pre]+1;
    if(l<r)
    {
        int mid=l+r>>1;
        if(x<=mid)L[rt]=update(L[pre],l,mid,x);
        else R[rt]=update(R[pre],mid+1,r,x);
    }
    return rt;
}
inline int query(int u,int v,int ql,int qr,int l,int r)
{
    if(ql<=l&&qr>=r)return sum[v]-sum[u];
    int mid=l+r>>1,ans=0;
    if(ql<=mid)ans+=query(L[u],L[v],ql,qr,l,mid);
    if(qr>mid)ans+=query(R[u],R[v],ql,qr,mid+1,r);
    return ans;
}
int main()
{
    int t;
    cin>>t;
    for(int W=1;W<=t;W++)
    {
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        T[0]=sum[0]=L[0]=R[0]=tot=0;
        for(int i=1;i<=n;i++)
        {
            T[i]=update(T[i-1],1,Max,a[i]);
        }
        int l,r,p,k,x=0;
        for(int i=1;i<=m;i++)
        {
            cin>>l>>r>>p>>k;
            l^=x;r^=x;p^=x;k^=x;
            int pl=0,pr=Max;
            while(pl<pr)
            {
                int mid=pl+pr>>1;
                if(query(T[l-1],T[r],max(1,p-mid),min(Max,p+mid),1,Max)>=k)
                {
                    x=mid;
                    pr=mid;
                }
                else pl=mid+1;
            }
            cout<<x<<endl;
        }
    }
    return 0;
}