2019 Multi-University Training Contest 1 1002 线性基

题意:序列操作1.末尾加数 2.求区间异或最大值。

 

求区间异或的最大值很容易就想到用线性基去做

比赛的时候试了分块和线段树去维护线性基,结果都TLE了

之前也想到了维护后缀线性基，但是不知道怎么处理区间，没想到是在线性基里维护出现的位置

正解应该是对每一个后缀维护一个线性基，每次插入元素的时候都在最高位选择出现位置偏右的数,查询的时候只要当前位的出现为止在l的右边就代表区间可以产生这个数

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
struct xj{
    int a[32],pos[32];
    inline void init(){
        for(int i = 0; i <= 30; i ++) a[i] = pos[i] = 0;
    }
    inline void add(int x,int p){
        for(int i = 30; i >= 0 ; i --){
            if(!(x & (1 << i))) continue;
            if(!a[i]){
                a[i] = x; pos[i] = p;
                break;
            }else{
                if(p > pos[i])swap(a[i],x),swap(p,pos[i]);
                x ^= a[i];
            }
        }
    }
    inline int getmax(int l){
        int ans = 0;
        for(int i = 30; i >= 0; i --){
            if(pos[i] < l) continue;
            if(ans < (ans ^ a[i])) ans ^= a[i];
        }
        return ans;
    }
}pre[maxn];
int main(){
    int T; Sca(T);
    while(T--){
        Sca2(N,M);
        for(int i = 1; i <= N ; i ++){
            int x = read();
            pre[i] = pre[i - 1];
            pre[i].add(x,i);
        }
        int ans = 0;
        while(M--){
            int op = read();
            if(!op){
                int l = read() ^ ans,r = read() ^ ans;
                l = l % N + 1; r = r % N + 1;
                if(l > r) swap(l,r);
                ans = pre[r].getmax(l);
                Pri(ans);
            }else{
                int x = read() ^ ans;
                N++;
                pre[N] = pre[N - 1];
                pre[N].add(x,N);
            }
        }
    } 
    return 0;
}