Codeforces Round #340 (Div. 2) E. XOR and Favorite Number （莫队）

题面

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.

Input
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.

Output
Print m lines, answer the queries in the order they appear in the input.

Examples
inputCopy
6 2 3
1 2 1 1 0 3
1 6
3 5
outputCopy
7
0
inputCopy
5 3 1
1 1 1 1 1
1 5
2 4
1 3
outputCopy
9
4
4

思路

没有修改只有询问，而且数据量是1e5，那么我们就可以考虑一下莫队了，对原数组做前缀和，然后离线询问，维护区间就行。

代码实现

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;};
void check_max (int &a,int b) { a=max (a,b);}
void check_min (int &a,int b) { a=min (a,b);}
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}

const int maxn=1<<20;
struct node {int l,r,id;}q[maxn];
ll flag[maxn];
int lim;
int a[maxn],i,l,r,k,n,m;
ll ans[maxn];
int pos[maxn];
int L=1,R=0;
ll now=0;
bool cmp (node &a,node &b) {return pos[a.l]==pos[b.l]?a.r<b.r:pos[a.l]<pos[b.l]; }


void add (int x) {
    now+=flag[a[x]^k];
    flag[a[x]]++;
}

void del (int x) {
    flag[a[x]]--;
    now-=flag[a[x]^k];
}

void solve () {
    scanf ("%d%d%d",&n,&m,&k);
    while (lim*lim<n) lim++;
    rep (i,1,n) pos[i]=(i-1)/lim+1;
    rep (i,1,n) {
        scanf ("%d",&a[i]);
        a[i]^=a[i-1];
    }
    rep (i,1,m) {
        scanf ("%d%d",&q[i].l,&q[i].r);
        q[i].id=i;
    }
    sort (q+1,q+1+m,cmp);
    flag[0]=1;
    for (i=l=1,r=0;i<=m;i++) {
        int L=q[i].l,R=q[i].r;
        if (r<R) {for (r++;r<=R;r++) add (r); r--;}
        if (r>R) {for (;r>R;r--) del (r);}
        if (l<L) for (;l<L;l++) del (l-1);
        if (l>L) {for (l--;l>=L;l--) add (l-1);l++;}
        ans[q[i].id]=now; 
    }
    rep (i,1,m) printf ("%lld
",ans[i]);
}

int main () {
    int t=1;
    while (t--) {
        solve ();
    }

    return 0;
}