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
题意：
m次询问，每次询问，问区间[l,r]之间|p-a[i]|第k小的数值是多少，然后得到的答案需要异或下一次所给的l,r,p,k得到真正的l，r，p ，k。
分析：这样的每一次异或就导致了，这题必须是强制在线的，但是一看这道题目竟然是15s的复杂度，那么可不可用线段树呐，每个节点存一个vector，然后排序，这样不就可以在查询的时候之间就找到了（下面的是错误代码QWQ）
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstring>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define rep(i,a,b) for(int i=a;i<=b;i++)
typedef long long ll;
using namespace std;
const int N=1e6+10;
const int INF=0x3f3f3f3f;
vector<int>ve[N<<2];
int a[N];
int n,m,prex;
void build(int id,int l,int r){
ve[id].clear();
for(int i=l;i<=r;i++)
ve[id].push_back(a[i]);
sort(ve[id].begin(),ve[id].end());
if(l==r)
return ;
int mid=(l+r)/2;
build(id<<1,l,mid);
build(id<<1|1,mid+1,r);
}
int query(int id,int l,int r,int L,int R,int k,int p){
if(L<=l && R>=r){
return abs(ve[id][k-1]-p);
}
int mid=(l+r)/2;
if(R<=mid)
return query(id<<1,l,mid,L,R,k,p);
else if(L>mid)
return query(id<<1|1,mid+1,r,L,R,k,p);
else{
return min(query(id<<1,l,mid,L,mid,k,p),query(id<<1|1,mid+1,r,mid+1,R,k,p));
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
int T;
scanf("%d",&T);
while(T--){prex=0;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
build(1,1,n);
for(int i=1;i<=m;i++){int l,r,p,k;
scanf("%d%d%d%d",&l,&r,&p,&k);
l^=prex;r^=prex;p^=prex;k^=prex;
printf("%d\n",prex=query(1,1,n,l,r,k,p));
}
}
return 0;
}
交上我就后悔了，思路不对啊，这样找的是个啥啊，但是…………TLE了，那怎么做呢，既然是有序的，可不可以二分呢；怎么二分，分谁呢？？？
分析题目中给出的式子： |p-a[i]|
是找所给的区间第K小的数是多少，那么不就可以查询|p-a[i]|在[l,r]的区间里面a[i]的个数有个不就好了，什么意思呢？就是每次我去二分|p-a[i]| 的绝对值，我们进一步化简|p-a[i]|=mid这个式子，就会得到
p-mid=a[i]=p+mid,那么就用线段树查询在数值[p-mid,p+mid]之间有多少个数不就可以二分了吗（线段树的操作实在是……tsl）
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstring>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define rep(i,a,b) for(int i=a;i<=b;i++)
typedef long long ll;
using namespace std;
const int N=1e6+10;
const int INF=0x3f3f3f3f;
vector<int>ve[N<<2];
vector<int>::iterator it;
int a[N];
int n,m,prex;
void build(int id,int l,int r){
ve[id].clear();
for(int i=l;i<=r;i++)
ve[id].push_back(a[i]);
sort(ve[id].begin(),ve[id].end());
if(l==r)
return ;
int mid=(l+r)/2;
build(id<<1,l,mid);
build(id<<1|1,mid+1,r);
}
int query(int id,int l,int r,int L,int R,int p){
if(L==l && R==r){
return upper_bound(ve[id].begin(),ve[id].end(),p)-ve[id].begin();
}
int mid=(l+r)/2;
if(R<=mid)
return query(id<<1,l,mid,L,R,p);
else if(L>mid)
return query(id<<1|1,mid+1,r,L,R,p);
else
return query(id<<1,l,mid,L,mid,p)+query(id<<1|1,mid+1,r,mid+1,R,p);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
int T;
scanf("%d",&T);
while(T--){prex=0;
scanf("%d%d",&n,&m);
int ub=-1,lb=0;
for(int i=1;i<=n;i++)
scanf("%d",&a[i]),ub=max(ub,a[i]);
build(1,1,n);
int rrrrr=ub;
for(int i=1;i<=m;i++){int l,r,p,k;
scanf("%d%d%d%d",&l,&r,&p,&k);
l^=prex;r^=prex;p^=prex;k^=prex;
ub=rrrrr;lb=0;
while(ub>=lb){
int mid=(lb+ub)/2;
int cnt=query(1,1,n,l,r,p+mid)-query(1,1,n,l,r,p-mid-1);
if(cnt<k)
lb=mid+1;
else
ub=mid-1;
}
prex=lb;
printf("%d\n",lb);
}
}
return 0;
}
接近九秒，QWQ，还有说用主席树来敲的，同样也是二分来实现的，自己也照着来了一发，也可以当模板了
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstring>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define rep(i,a,b) for(int i=a;i<=b;i++)
typedef long long ll;
using namespace std;
const int N=1e6+10;
const int INF=0x3f3f3f3f;
int a[N],root[N];
int n,m,prex,cnt;
struct node{
int sum,l,r;
}tr[N*30];
void update(int l,int r,int &now,int last,int pos){
tr[++cnt]=tr[last];
now=cnt;
tr[now].sum++;
if(l==r) return ;
int mid=(l+r)/2;
if(pos<=mid)
update(l,mid,tr[now].l,tr[last].l,pos);
else
update(mid+1,r,tr[now].r,tr[last].r,pos);
}
int query(int L,int R,int l,int r,int x,int y){
if(L<=l&& r<=R)
return tr[y].sum-tr[x].sum;
int mid=(l+r)/2;
// int ans=0;
// if(L<=mid) ans+=query(L,R,l,mid,tr[x].l,tr[y].l);
// if(R>mid) ans+=query(L,R,mid+1,r,tr[x].r,tr[y].r);
// return ans;
if(R<=mid) return query(L,R,l,mid,tr[x].l,tr[y].l);
else if(L>mid) return query(L,R,mid+1,r,tr[x].r,tr[y].r);
else return query(L,mid,l,mid,tr[x].l,tr[y].l)+query(mid+1,R,mid+1,r,tr[x].r,tr[y].r);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
int T;
scanf("%d",&T);
while(T--){prex=0;cnt=0;
scanf("%d%d",&n,&m);
memset(tr,0,sizeof tr);
memset(root,0,sizeof root);
int ub=-1,lb=0;
for(int i=1;i<=n;i++)
scanf("%d",&a[i]),ub=max(ub,a[i]);
for(int i=1;i<=n;i++)
update(1,ub,root[i],root[i-1],a[i]);
int maxv=ub;
for(int i=1;i<=m;i++){int l,r,p,k;
scanf("%d%d%d%d",&l,&r,&p,&k);
l^=prex;r^=prex;p^=prex;k^=prex;
ub=1e6;lb=0;
while(ub>=lb){
int mid=(lb+ub)/2;
int cnt=query(max(1,p-mid),min(p+mid,maxv),1,maxv,root[l-1],root[r]);
if(cnt<k)
lb=mid+1;
else
ub=mid-1;
}
prex=lb;
printf("%d\n",lb);
}
}
return 0;
}
两种查询，发现第一种查询会快一点；
