题意
Sol
一开始以为K每次都是给出的想了半天不会做。
然而发现读错题了维护个前缀异或和然后直接莫队搞就行，。
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, mod = 998244353, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '
';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
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;
}
int N, Q, K, a[MAXN], num[MAXN], belong[MAXN], block;
LL now, ans[MAXN];
struct Query {
int l, r, id;
bool operator < (const Query &rhs) const {
return belong[l] == belong[rhs.l] ? r < rhs.r : belong[l] < belong[rhs.l];
}
}q[MAXN];
void Add(int x) {
now += num[K ^ x];
num[x]++;
}
void Delet(int x) {
num[x]--;
now -= num[K ^ x];
}
void solve() {
int l = 1, r = 0;
for(int i = 1; i <= Q; i++) {
while(r < q[i].r) Add(a[++r]);
while(r > q[i].r) Delet(a[r--]);
while(l < q[i].l) Delet(a[l++]);
while(l > q[i].l) Add(a[--l]);
ans[q[i].id] = now;
}
}
signed main() {
N = read(); Q = read(); K = read(); block = sqrt(N);
for(int i = 1; i <= N; i++) a[i] = read() ^ a[i - 1], belong[i] = (i - 1) / block + 1;
for(int i = 1; i <= Q; i++) q[i].l = read() - 1, q[i].r = read(), q[i].id = i;
sort(q + 1, q + Q + 1);
solve();
for(int i = 1; i <= Q; i++) cout << ans[i] << '
';
return 0;
}