嘟嘟嘟


前缀和+莫队。


先用前缀和预处理异或，于是问题变成了在\([L - 1, R]\)中求两个数异或等于\(k\)的数对个数。


然后就离线排序，按套路维护两个指针加加减减，并维护一个桶，每一次加\(x\)，答案就加上\(bac[x ^ k]\)，并且\(++bac[x]\)，删除就减去贡献。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, k, S, sum[maxn];
struct Node
{
  int L, R, id, b;
  In bool operator < (const Node& oth)const
  {
    return b < oth.b || (b == oth.b && R < oth.R);
  }
}q[maxn];

ll cnt = 0, bac[maxn], ans[maxn];
In void add(int x)
{
  cnt += bac[x ^ k], ++bac[x];
}
In void del(int x)
{
  --bac[x], cnt -= bac[x ^ k];
}

int main()
{
  n = read(), m = read(), k = read(); S = sqrt(n);
  for(int i = 1; i <= n; ++i) sum[i] = read(), sum[i] ^= sum[i - 1];
  for(int i = 1; i <= m; ++i)
    {
      int L = read() - 1, R = read();
      q[i] = (Node){L, R, i, (L - 1) / S + 1};
    }
  sort(q + 1, q + m + 1);
  for(int i = 1, l = 1, r = 0; i <= m; ++i)
    {
      while(l < q[i].L) del(sum[l++]);
      while(l > q[i].L) add(sum[--l]);
      while(r < q[i].R) add(sum[++r]);
      while(r > q[i].R) del(sum[r--]);
      ans[q[i].id] = cnt;
    }
  for(int i = 1; i <= m; ++i) write(ans[i]), enter;
  return 0;
}