[CQOI2018]异或序列

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;

}

巴特西

[CQOI2018]异或序列

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