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

Source

2019 Multi-University Training Contest 4

题目大意：T组样例，N个数 M次询问，L,R,P，K  问你L,R区间内 距离P第K近的距离是多少

思路：比赛的时候没啥思路，没写，看题解发现，主席树可以求区间内<=K的数有多少个  但是我们要求的是与某个值相减绝对值第K小啊，这有什么关系吗？

假设我们知道答案是几，那么此时求出来L,R区间内的数与P相减第K小是不是就是我们这个答案呢？ 显然是的，但是问题是我们不知道答案啊

这就是二分的思想了，二分答案，查询范围在P-ans，P+ans之间的数有多少个，如果个数大于等于K的话，证明我们的ans还可以减小，一直这样下去

就会得到正确的答案了

看代码：

#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn=1e5+5;
int N,M;
int sz,cnt=0,sum=0;
int root[maxn],a[maxn],v[maxn];
int L,R,P,K,X=0;
struct Node
{int l,r,sum;Node(){l=r=sum=0;}
}T[maxn<<6];
void Init()
{cnt=0;root[0]=0;T[cnt].l=T[cnt].r=T[cnt].sum=0;X=0;
}
void Read(int &n)//只能读入整数
{n=0;int f=1;//用于记录正负char ch=getchar();while(ch<'0'||ch>'9')//
    {if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9')//下面两种用法都可以
    {//n=(n<<1)+(n<<3)+(ch-'0');//
        n=(n<<1)+(n<<3)+(ch^48);//相当于  x*10+ch-'0';ch=getchar();}n=n*f;//切记乘以f  （记录的是+ -）return ;
}
void Update(int l,int r,int &x,int y,int pos)
{T[++cnt]=T[y];T[cnt].sum++;x=cnt;if(l==r) return ;int mid=(l+r)>>1;if(pos<=mid) Update(l,mid,T[x].l,T[y].l,pos);else Update(mid+1,r,T[x].r,T[y].r,pos);
}
int getid(int x)
{return lower_bound(v+1,v+1+sz,x)-(v);
}
void Query(int l,int r,int x,int y,int h)
{
//    cout<<"l:"<<l<<" r:"<<r<<" sum:"<<sum<<endl;if(l==r){if(h>=v[l]) sum+=T[x].sum-T[y].sum;return ;}int mid=(l+r)>>1;if(h<=v[mid]) Query(l,mid,T[x].l,T[y].l,h);else{sum+=T[T[x].l].sum-T[T[y].l].sum;Query(mid+1,r,T[x].r,T[y].r,h);}
}
bool judge(int x)
{sum=0;Query(1,sz,root[R],root[L-1],P+x);int sum1=sum;sum=0;Query(1,sz,root[R],root[L-1],P-x-1);int sum2=sum;return sum1-sum2>=K;
}
int main()
{int T;
//    scanf("%d",&T);
Read(T);int ca=1;while(T--){Init();
//        scanf("%d%d",&N,&M);
Read(N);Read(M);for(int i=1;i<=N;i++){scanf("%d",&a[i]);v[i]=a[i];}sort(v+1,v+1+N);sz=unique(v+1,v+1+N)-(v+1);
//        for(int i=1;i<=sz;i++) cout<<v[i]<<" ";cout<<endl;
//        for(int i=1;i<=N;i++) cout<<getid(a[i])<<" ";cout<<endl;for(int i=1;i<=N;i++) Update(1,sz,root[i],root[i-1],getid(a[i]));while(M--){sum=0;
//            scanf("%d%d%d%d",&L,&R,&P,&K);
Read(L);Read(R);Read(P);Read(K);L^=X;R^=X;P^=X;K^=X;int l=0,r=1e6,ans;while(l<=r)//二分答案
            {int mid=(l+r)>>1;if(judge(mid)){ans=mid;r=mid-1;}else l=mid+1;}printf("%d\n",ans);X=ans;}}return 0;
}