看完题，容易想到把求一段区间改成求 前r的异或和 ^ 前l-1的异或和 等于k， 那么这个东西发现可以直接用莫队统计，没了. c++代码如下：
#include<bits/stdc++.h> #define rep(i,x,y) for(register int i = x ; i <= y ; ++ i) #define repd(i,x,y) for(register int i = x ; i >= y ; -- i) using namespace std; typedef long long ll; template<typename T>inline void read(T&x) { char c;int sign = 1;x = 0; do { c = getchar(); if(c == '-') sign = -1; }while(!isdigit(c)); do { x = x * 10 + c - '0'; c = getchar(); }while(isdigit(c)); x *= sign; } const int N = 1e5+850; int n,m,k,a[N]; int num[N],ans[N],belong[N]; struct DATA { int l,r,id; }q[N]; const bool cmp(DATA a,DATA b) { return belong[a.l] < belong[b.l] || belong[a.l] == belong[b.l] && belong[a.r] < belong[b.r]; } int main() { read(n); read(m); read(k); rep(i,1,n) read(a[i]); rep(i,2,n) a[i] = a[i]^a[i-1]; rep(i,1,m) read(q[i].l), read(q[i].r),q[i].id = i; int d = max(1,(int)pow(n,2.0/3)); rep(i,1,n) belong[i] = i/d; sort(q + 1,q + 1 + m,cmp); int r = 0,l = 1,now = 0; rep(i,1,m) { --q[i].l; while(l < q[i].l) { --num[a[l]]; now -= num[a[l++]^k]; } while(l > q[i].l) { now += num[a[--l]^k]; ++num[a[l]]; } while(r < q[i].r) { now += num[a[++r]^k]; ++num[a[r]]; } while(r > q[i].r) { --num[a[r]]; now -= num[a[r--]^k]; } ans[q[i].id] = now; } rep(i,1,m) printf("%d\n",ans[i]); return 0; }