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Codeforces Round #340 (Div. 2) E. XOR and Favorite Number 莫队算法
Bob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query is given by a pair li and ri and asks you to count the number of pairs of integers i and j, such that l ≤ i ≤ j ≤ r and the xor of the numbers ai, ai + 1, ..., aj is equal to k.
The first line of the input contains integers n, m and k (1 ≤ n, m ≤ 100 000, 0 ≤ k ≤ 1 000 000) — the length of the array, the number of queries and Bob‘s favorite number respectively.
The second line contains n integers ai (0 ≤ ai ≤ 1 000 000) — Bob‘s array.
Then m lines follow. The i-th line contains integers li and ri (1 ≤ li ≤ ri ≤ n) — the parameters of the i-th query.
Print m lines, answer the queries in the order they appear in the input.
6 2 3
1 2 1 1 0 3
1 6
3 5
7
0
5 3 1
1 1 1 1 1
1 5
2 4
1 3
9
4
4
In the first sample the suitable pairs of i and j for the first query are: (1, 2), (1, 4), (1, 5), (2, 3), (3, 6), (5, 6), (6, 6). Not a single of these pairs is suitable for the second query.
In the second sample xor equals 1 for all subarrays of an odd length.
题意：找出区间有多少对异或和为k的对数；
思路：莫队算法，主要是如何o（1）更新，a[l]^a[l+1].....^a[r]=pre[r]^pre[l-1]=k;pre[i]表示a[1]^a[2]^....^a[i];
pos[r]^pre[l-1]=k; pre[r]=pre[l-1]^k;更新pre；
#include<bits/stdc++.h>using namespace std;#define ll long long#define pi (4*atan(1.0))const int N=1e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7;const ll INF=1e18+10;int si[N];struct is{ int l,r,pos; bool operator < (const is &b)const { if(si[l]==si[b.l]) return r<b.r; return si[l]<si[b.l]; }}q[N];int a[N];int n,m,k;int l,r;ll out[N],ans;int flag[M<<1];void add(int pos){ ans+=flag[k^a[pos]]; flag[a[pos]]++;}void del(int pos){ flag[a[pos]]--; ans-=flag[k^a[pos]];}int main(){ scanf("%d%d%d",&n,&m,&k); int kuai=sqrt(n); for(int i=1;i<=n;i++) scanf("%d",&a[i]),si[i]=(i-1)/kuai+1,a[i]^=a[i-1]; for(int i=1;i<=m;i++) scanf("%d%d",&q[i].l,&q[i].r),q[i].pos=i; sort(q+1,q+m+1); l=1; r=0; ans=0; flag[0]=1; for(int i=1;i<=m;i++) { while(l<q[i].l) { del(l-1); l++; } while(l>q[i].l) { l--; add(l-1); } while(r<q[i].r) { r++; add(r); } while(r>q[i].r) { del(r); r--; } out[q[i].pos]=ans; } for(int i=1;i<=m;i++) printf("%lld\n",out[i]); return 0;}
Codeforces Round #340 (Div. 2) E. XOR and Favorite Number 莫队算法