Codeforces617E XOR and Favorite Number（分块异或）

Bob has a favorite number k and a_i of length n. Now he asks you to answer m queries. Each query is given by a pair l_i and r_i 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 a_i, a_i + 1, ..., a_jis 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 a_i (0 ≤ a_i ≤ 1 000 000) — Bob's array.

Then m lines follow. The i-th line contains integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the parameters of the i-th query.

Output

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

Example

Input

6 2 3
1 2 1 1 0 3
1 6
3 5

Output

7
0

Input

Output

9
4
4

Note

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：

给定数列a[],和m个询问 Q(L,R),回答每个询问中有多少对(L<=i<=j<=R) ，使得异或为k。

异或转化为前缀和处理。然后就差不多交给分块处理了。

离线处理里比较好理解的一种，高中就会了，注意这里不多BB了。

#include<cstdio>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cmath>

#define ll long long

using namespace std;

const int maxn=;

int a[maxn],pre[maxn],num[<<],n,m,k,sqrtn;

struct Query{

    int id,l,r;  ll ans;

}q[maxn];

bool cmp(const Query a,const Query b)

{

    if(a.l/sqrtn==b.l/sqrtn) return a.r<b.r;

    return a.l<b.l;

}

bool cmp2(const Query a,const Query b)

{

    return a.id<b.id;

}

int main()

{

     scanf("%d%d%d",&n,&m,&k);

     sqrtn=(int)sqrt(n);

     for(int i=;i<=n;i++) {

          scanf("%d",&a[i]);

          pre[i]=pre[i-]^a[i];

     }

     for(int i=;i<m;i++){

         scanf("%d%d",&q[i].l,&q[i].r);

         q[i].id=i;

     }

     sort(q,q+m,cmp);

     int l=,r=;

     num[pre[]]++;num[]++;

     ll cur=(a[]==k?:);

     for(int i=;i<m;i++)

     {

        while(r<q[i].r){

            cur+=num[pre[r+]^k];

            r++;

            num[pre[r]]++;

        }

        while(l<q[i].l){

            num[pre[l-]]--;

            cur-=num[pre[l-]^k];

            l++;

        }

        while(l>q[i].l){

            cur+=num[pre[l-]^k];

            num[pre[l-]]++;

            l--;

        }

        while(r>q[i].r){

            num[pre[r]]--;

            cur-=num[pre[r]^k];

            r--;

        }

        q[i].ans=cur;

     }

     sort(q,q+m,cmp2);

     for(int i=;i<m;i++) printf("%lld\n",q[i].ans);

     return ;

}

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