【Codeforces 1100 F Ivan and Burgers 】线性基

F

给你n个数 q次询问 问你这q次区间内能异或的最大值

异或的最大值我们容易想到线性基 但是要区间内 我们不能暴力扫区间构造

所以我们把问题离线了再去做 将询问按照 r 的升序排序 那么按顺序插入 1 - r 的数目构造线性基 然后记录一下pos 我们每个线性基优先取右边的 这样查询的时候只要在pos >= l的位置查询就行 因为线性基只有20组 所以可以暴力查询 也就是log级别的插入与查询

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
const int MAX_N = 500025;
int a[25],pos[25],arr[MAX_N],ans[MAX_N];
struct node
{
    int l,r,id;
    bool operator< (const node other) const
    {
        if(r==other.r) return l < other.l;
        return r < other.r;
    }
}q[MAX_N];
bool Insert(int val,int p)
{
    int u = val;
    for(int i = 20;i>=0;i--)
    {
        if(u&(1ll<<i))
        {
            if(!a[i])
            {
                a[i] = u;
                pos[i] = p;
                break;
            }
            if(pos[i]<p)
            {
                swap(a[i],u);
                swap(pos[i],p);
            }
            u^=a[i];
        }
    }
    return u>0;
}
void query(int l,int id)
{
    int ret = 0;
    for(int i = 20;i>=0;--i)
    {
        if((ret^a[i])>ret&&pos[i]>=l)
            ret^=a[i];
    }
    ans[id] = ret;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,Q,r = 1;
    scanf("%d",&n);
    for(int i = 1;i<=n;++i)
        scanf("%d",&arr[i]);
    scanf("%d",&Q);
    for(int i = 1;i<=Q;++i)
    {
        scanf("%d%d",&q[i].l,&q[i].r);
        q[i].id = i;
    }
    sort(q+1,q+1+Q);
    for(int i = 1;i<=Q;++i)
    {
        while(r<=q[i].r)
        {
            Insert(arr[r],r);
            r++;
        }
        query(q[i].l,q[i].id);
    }
    for(int i = 1;i<=Q;++i)
        printf("%d\n",ans[i]);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}