Magic-yl linear change
Online practices mandatory, the need to maintain a linear base prefix, every time the number of new entrants, we should mention the right by the number of high linear base
After linear base such maintenance at the time of the query interval XOR and, simply put r linear base prefix appears far greater than l and the number of exclusive-OR and XOR up more on the line, the new routine! !
#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;
}