2019 Multi-University Training Contest 1 - Operation

魔改线性基

强制在线的做法，需要维护一个前缀的线性基，每次新加入数的时候，要把靠右边的数提到线性基的高位

这样维护的线性基，在查询区间异或和的时候，只需要把r为前缀的线性基出现为止大于l且异或之后和更大的数异或起来就行了，新套路！！

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        ret = (ret << 3) + (ret << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -ret : ret;
}
inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 1000005;
int _, n, m, pos[N][32], l, r;
LL b[N][32], val;

int main(){

    for(scanf("%d", &_); _; _ --){
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i ++){
            scanf("%lld", &val);
            for(int j = 0; j <= 31; j ++)
                b[i][j] = b[i - 1][j], pos[i][j] = pos[i - 1][j];
            int cur = i;
            for(int j = 31; j >= 0; j --){
                if(val & (1LL << j)){
                    if(!b[i][j]){
                        b[i][j] = val, pos[i][j] = cur;
                        break;
                    }
                    else{
                        if(pos[i][j] < cur) swap(pos[i][j], cur), swap(b[i][j], val);
                        val ^= b[i][j];
                    }
                }
            }
        }
        LL opt, l, r, cnt = n;
        LL val, ret = 0;
        while(m --){
            scanf("%lld", &opt);
            if(opt == 0){
                scanf("%lld%lld", &l, &r);
                l = (l ^ ret) % cnt + 1, r = (r ^ ret) % cnt + 1;
                if(l > r) swap(l, r);
                ret = 0;
                for(int i = 31; i >= 0; i --){
                    if(pos[r][i] >= l && (ret ^ b[r][i]) > ret)
                        ret ^= b[r][i];
                }
                printf("%lld\n", ret);
            }
            else{
                scanf("%lld", &val);
                val ^= ret, cnt ++;
                for(int i = 0; i <= 31; i ++)
                    b[cnt][i] = b[cnt - 1][i], pos[cnt][i] = pos[cnt - 1][i];
                int cur = cnt;
                for(int i = 31; i >= 0; i --){
                    if(val & (1LL << i)){
                        if(!b[cnt][i]){
                            b[cnt][i] = val, pos[cnt][i] = cur;
                            break;
                        }
                        else{
                            if(pos[cnt][i] < cur) swap(pos[cnt][i], cur), swap(b[cnt][i], val);
                            val ^= b[cnt][i];
                        }
                    }
                }
            }
        }
        for (int i=1;i<=n;++i)
            for (int j=30;j>=0;--j) b[i][j] = pos[i][j] =0;
    }
    return 0;
}