很显然每个数的每一位最多只会修改一遍。于是拆位，每一位开个并查集，存下一个不拥有这一位的数，就可以暴力修改了。
但是空间是 \(O(n \log V)\) 的，炸了。于是可以考虑手写 i24 类，同时并查集寻找祖先不要用递归版的路径压缩，然后就过了。
code
// Problem: P8955 「VUSC」Card Tricks
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P8955
// Memory Limit: 100 MB
// Time Limit: 3000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
bool Mst;
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 20], *p1 = buf, *p2 = buf;
inline int read() {
char c = getchar();
int x = 0;
for (; !isdigit(c); c = getchar()) ;
for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x;
}
const int maxn = 1000100;
int n, m, K, a[maxn], ans[maxn];
struct i24 {
unsigned char a, b, c;
inline void set(int x) {
a = x & 255;
b = (x >> 8) & 255;
c = (x >> 16) & 255;
}
inline int val() {
return a + ((int)b << 8) + ((int)c << 16);
}
} fa[30][maxn];
inline int find(int k, int x) {
while (fa[k][x].val() != x) {
int t = fa[k][fa[k][x].val()].val();
fa[k][x].set(t);
x = t;
}
return x;
}
void solve() {
n = read();
m = read();
K = read();
for (int i = 1; i <= n; ++i) {
a[i] = read();
ans[i] = -1;
}
for (int j = 0; j < 30; ++j) {
for (int i = 1; i <= n + 1; ++i) {
fa[j][i].set((a[i] & (1 << j)) ? i + 1 : i);
}
}
for (int t = 1, l, r, x; t <= m; ++t) {
l = read();
r = read();
x = read();
for (int j = 0; j < 30; ++j) {
if ((~x) & (1 << j)) {
continue;
}
for (int i = find(j, l); i <= r; i = find(j, i)) {
a[i] |= (1 << j);
if (a[i] > K && ans[i] == -1) {
ans[i] = t;
}
fa[j][i].set(i + 1);
}
}
}
for (int i = 1; i <= n; ++i) {
printf("%d ", ans[i]);
}
}
bool Med;
int main() {
fprintf(stderr, "%.2lf MB\n", (&Mst - &Med) / 1048576.);
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}