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;
}
hdu6621 二分加主席树
原文：https://www.cnblogs.com/chen99/p/11294015.html
