Codeforces Round #340 (Div. 2) E 莫队+前缀异或和

时间：2023-03-10 04:46:14
Codeforces Round #340 (Div. 2) E 莫队+前缀异或和
E. XOR and Favorite Number
time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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
Input
6 2 3
1 2 1 1 0 3
1 6
3 5
Output
7
0
Input
5 3 1
1 1 1 1 1
1 5
2 4
1 3
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：给你一个长度为n的序列 q个查询[l,r]  问[l,r] 有多少个子区间的异或和为k

题解：莫队+前缀异或和

 #include<iostream>

 #include<cstdio>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #include<map>

 #include<queue>

 #include<stack>

 #include<vector>

 #include<set>

 #define ll __int64

 using namespace std;

 int n,m,k;

 struct node

 {

     int l,r,id;

 }N[];

 int p[];

 int block;

 int a[];

 int x[];

 int mp[];

 ll ans=;

 ll re[];

 int cmp(struct node aa,struct node bb)

 {

     if(p[aa.l]==p[bb.l])

         return  aa.r<bb.r;

     else

         return p[aa.l]<p[bb.l];

 }

 void update(int w,int h)

 {

     if(h==){

         ans=ans+mp[x[w]^k];

         mp[x[w]]++;

     }

     else

     {

         mp[x[w]]--;

         ans=ans-mp[x[w]^k];

     }

 }

 int main()

 {

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

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

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

     x[]=;

     mp[]=;

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

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

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

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

         N[i].id=i;

     }

     block=(int)sqrt((double)n);

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

         p[i]=(i-)/block+;

     sort(N+,N++m,cmp);

     ans=;

     for(int i=,l=,r=;i<=m;i++)

     {

         for(;r<N[i].r;r++) update(r+,);

         for(;l>N[i].l;l--) update(l-,);// 取异或的原因

         for(;r>N[i].r;r--) update(r,-);

         for(;l<N[i].l;l++) update(l-,-);//

         re[N[i].id]=ans;

     }

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

     printf("%I64d\n",re[i]);

    return ;

 }

 /*

 10 2 0

 0 0 0 0 0 0 0 0 0 0

 2 8

 2 8

 */

相关文章