I. Five Day Couple--"Today's Toutiao Cup" The First Hubei University Programming Contest (Network Synchronized Contest)

Topic description: link here

The github address of this set of questions (which contains data, question solutions, and on-site rankings): click here

Link: https://www.nowcoder.com/acm/contest/104/H
Source: Niuke.com

Topic description

Mingming, a cute girl of ACM/ICPC team of Wuhan University, is alone since graduate from high school. Last year, she used a program to match boys and girls who took part in an active called Boy or Girl friend in five days.

She numbered n () boys from 1 to \(n\), by their date of birth, and given i-th boy a number () in almost random. (We do not mean that in your input is generated in random.). Then she numbered m () girls from 1 to m, and given i-th girl a number () in the same way.

Also, i-th girl said that she only wanted to be matched to a boy whose age is between , which means that she should only be matched to a boy numbered from  , ().

Mingming defined a rate R(i,j) to measure the score when the i-th boy and j-th girl matched. Where  where means bitwise exclusive or. The higher, the better.

Now, for every girl, Mingming wants to know the best matched boy, or her "Mr. Right" can be found while her . As this is the first stage of matching process and Mingming will change the result manually, two girls can have the same "Mr. Right".

Enter description:

The first line contains one number n.

The second line contains n integers, the i-th one is .

The third line contains an integer m.

Then followed by m lines, the j-th line contains three integers .

Output description:

Output m lines, the i-th line contains one integer, which is the matching rate of i-th girl and her Mr. Right.

Example 1

enter

4
19 19 8 10
2
1 1 4
5 1 4

output

18
22 

question meaning: It is to give you n numbers, and then ask m times, each time there are three numbers b, l, r, find the value within the range of [l, r], and the value with the largest xor value of b is the largest 

. Use a dictionary 

but this involves interval query, so a persistent dictionary is required

/*
    data:2018.04.27
    author:gsw
    link:https://www.nowcoder.com/acm/contest/104/H
    account:tonygsw
*/
#define ll long long
#define IO ios::sync_with_stdio(false);

#include<iostream>
#include<math.h>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=1e5+5;

class Node{
    public:
        int cnt,ls,rs;
};
Node tr[maxn*40];
int root[maxn];int cnt;

int  in ( int pre, int x, int deep)
{
    int num == cnt;
    tr [num] = tr [pre];
    tr[num].cnt=tr[pre].cnt+1;
    if(deep<0)return num;
    if(!((x>>deep)&1))tr[num].ls=in(tr[pre].ls,x,deep-1);
    else tr[num].rs=in(tr[pre].rs,x,deep-1);
    return num;
}
int query(int l,int r,int x,int deep)
{
    if(deep<0)return 0;
    if(!((x>>deep)&1))
    {
        if(tr[tr[r].rs].cnt>tr[tr[l].rs].cnt)return (1<<deep)+query(tr[l].rs,tr[r].rs,x,deep-1);
        else return query(tr[l].ls,tr[r].ls,x,deep-1);
    }
    else
    {
        if(tr[tr[r].ls].cnt>tr[tr[l].ls].cnt)return (1<<deep)+query(tr[l].ls,tr[r].ls,x,deep-1);
        else return query(tr[l].rs,tr[r].rs,x,deep-1);
    }
}


intmain ()
{
    int n,x,m,b,l,r;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        root[i]=in(root[i-1],x,29);
    }
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&b,&l,&r);
        printf("%d\n",query(root[l-1],root[r],b,29));
    }
}