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

E. XOR and Favorite Number

题目连接：

http://www.codeforces.com/contest/617/problem/E

Descriptionww.co

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.

Sample Input

6 2 3

1 2 1 1 0 3

1 6

3 5

Sample Output

7

0

Hint

题意

给你n个数，然后M次询问，问你l,r区间内有多少对数，使得a[i]^a[j] = k

题解：

无修改，而且可以知道[l,r]可以O(1)就出[l-1,r],[l,r+1],[l+1,r],[l,r-1]的数据的

所有很显然的莫队算法搞一搞就好了

直接大暴力，注意不能再带log，所以直接开数组存就好了

注意，数组得开大一点哦

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 120010;

int a[maxn],pos[maxn];
long long ans,flag[5000000];
long long Ans[maxn];
int k;
struct query
{
    int l,r,id;
}Q[maxn];
bool cmp(query a,query b)
{
    if(pos[a.l]==pos[b.l])
        return a.r<b.r;
    return pos[a.l]<pos[b.l];
}
void Updata(int x)
{
    ans+=flag[a[x]^k];
    flag[a[x]]++;
}
void Delete(int x)
{
    flag[a[x]]--;
    ans-=flag[a[x]^k];
}
int main()
{
    int n,m;
    scanf("%d%d%d",&n,&m,&k);
    int sz =ceil(sqrt(1.0*n));
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        pos[i]=(i-1)/sz;
    }
    for(int i=1;i<=n;i++)
        a[i]^=a[i-1];
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&Q[i].l,&Q[i].r);
        Q[i].id = i;
    }
    sort(Q+1,Q+1+m,cmp);
    int l = 1,r = 0;
    ans=0;
    flag[0]=1;
    for(int i=1;i<=m;i++)
    {
        int id = Q[i].id;
        while(r<Q[i].r)
        {
            r++;
            Updata(r);
        }
        while(l>Q[i].l)
        {
            l--;
            Updata(l-1);
        }
        while(r>Q[i].r)
        {
            Delete(r);
            r--;
        }
        while(l<Q[i].l)
        {
            Delete(l-1);
            l++;
        }
        Ans[id]=ans;
    }
    for(int i=1;i<=m;i++)
        printf("%lld\n",Ans[i]);
}