Codeforces Round #532(Div.2) - F. Ivan and Burgers - 선형 베이스

Ivan loves burgers and spending money. There are   burger joints on the street where Ivan
lives. Ivan has   friends, and the  -th friend suggested to meet at the joint li and walk
to the joint ri(li<ri). While strolling with the  i-th friend Ivan can visit all joints 
which satisfy li<xi<ri.
    
For each joint Ivan knows the cost of the most expensive burger in it, it costs     
burles. Ivan wants to visit some subset of joints on his way, in each of them he will buy
the most expensive burger and spend the most money. But there is a small issue: his card 
broke and instead of charging him for purchases, the amount of money on it changes as 
follows.
    
If Ivan had   burles before the purchase and he spent   burles at the joint, then after 
the purchase he would have   burles, where   denotes the bitwise XOR operation.
    
Currently Ivan has       burles and he wants to go out for a walk. Help him to determine 
the maximal amount of burles he can spend if he goes for a walk with the friend  . The 
amount of burles he spends is defined as the difference between the initial amount on his
account and the final account.

The first line contains one integer n  (1<=n<=500000) — the number of burger shops.
    
The next line contains n integers c1,c2,……,cn(0<=ci<=1e6), where ci  — the cost of the 
most expensive burger in the burger joint  .
    
The third line contains one integer q (1<=q<=500000) — the number of Ivan's friends.

Each of the next   lines contain two integers li and ri (1<=li<=ri<=n) — pairs of numbers
of burger shops between which Ivan will walk. 

Output q  lines,i-th of which containing the maximum amount of money Ivan can spend with 
the friend i .

4
7 2 3 4
3
1 4
2 3
1 3

7
3
7

5
12 14 23 13 7
15
1 1
1 2
1 3
1 4
1 5
2 2
2 3
2 4
2 5
3 3
3 4
3 5
4 4
4 5
5 5

12
14
27
27
31
14
25
26
30
23
26
29
13
13
7

  n  a1,a2,……,an，   q   ，          [l,r]，               ，    
   。

   
               ，   ，               ，  a[i]  ，i  ，     ， 
         。                i   。
             ，                i      l。

/*
 * @Author: Simon 
 * @Date: 2019-03-06 20:00:45 
 * @Last Modified by: Simon
 * @Last Modified time: 2019-03-06 20:26:44
 */
#include
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 500005
int a[25],v[maxn],pos[25]; 
struct node{
    int u,v,ord;
    bool operator <(const node a)const {
        if(v==a.v) return u<a.u;
        return v<a.v;
    }
}c[maxn];
void insert(int val,int p){//      
    for(int i=20;~i;i--){
        if(val>>i&1LL){
            if(!a[i]){
                a[i]=val;pos[i]=p;break;//    
            }
            if(p>pos[i]) swap(pos[i],p),swap(a[i],val); //p  ，    
            val^=a[i];
        }
    }
}
int query(int l){ //     
    int ans=0;
    for(int i=20;~i;i--){
        if(pos[i]>=l&&(ans^a[i])>ans) ans^=a[i]; //          [l,r] 
    }
    return ans;
}
int ans[maxn];
Int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;scanf("%lld",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",v+i);
    }
    int q;scanf("%lld",&q);
    for(int i=1;i<=q;i++) scanf("%lld%lld",&c[i].u,&c[i].v),c[i].ord=i;
    sort(c+1,c+q+1);int r=1;
    for(int i=1;i<=q;i++){
        while(r<=c[i].v){
            insert(v[r],r);r++;
        }
        ans[c[i].ord]=query(c[i].u);
    }
    for(int i=1;i<=q;i++) printf("%lld
",ans[i]);
    cin.get(),cin.get();
    return 0;
}

/*
 * @Author: Simon 
 * @Date: 2019-07-23 12:15:12 
 * @Last Modified by: Simon
 * @Last Modified time: 2019-07-23 13:39:15
 */
#include
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 500005
int a[maxn],base[maxn][25]/*     i     */,pos[maxn][25];
void insert(int val,int p){
    int k=p;
    for(int i=0;i<=20;i++) base[p][i]=base[p-1][i],pos[p][i]=pos[p-1][i]; //       p-1    ，       
    for(int i=20;i>=0;i--) if(val>>i&1){
        if(!base[p][i]){
            base[p][i]=val;
            pos[p][i]=k;
            break;
        }
        if (k > pos[p][i]) {
            swap(pos[p][i], k); //      ,        k，    p
            swap(base[p][i], val);
        }
        val ^= base[p][i];
    }
}
Int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i],insert(a[i],i);
    int q;cin>>q;
    while(q--){
        int l,r;cin>>l>>r;
        int ans=0;
        for(int i=20;i>=0;i--){
            if((ans^base[r][i])>ans&&pos[r][i]>=l) ans^=base[r][i];
        }
        cout<<ans<<endl;
    }
    cin.get(),cin.get();
    return 0;
}