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;
}