Any door: Http://codeforces.com/problemset/problem/617/E

E. XOR and favorite number

**Time ****limit per test**
**4 seconds**
**memory limit per test**
**Megabytes**
**input**
**Standard ****Input**
**Output**
**Standard ****Output**
Bob has a favorite number*k*and*a* *i* of length *n*. Now he asks your to answer *m* queries. Each query was given by a pair *l**i* and *R**i* and asks you to count th E 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**J* 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 *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.

Examplesinput

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 is: (1, 2), ( 1, 4), (1, 5), ( 2,3 ), (3, 6), (5, 6), (6, 6). Not a single of these pairs are suitable for the second query.

In the second, the sample XOR equals 1 for any subarrays of an odd length.

Main topic:

There is a series of N of length, M query (L,R) within how many pairs (i, j) make Ai ^ ai+1 ^ ... ^ aj = K;

Problem Solving Ideas:

The interval query here is offline and can be used in the legendary MO team algorithm (elegant and gorgeous brute force algorithm).

One of the most ingenious parts of an XOR topic is the use of A^b^a = B as a property.

This topic we have to preprocess the sum (i) before I number of XOR and, then ai ^ ai+1 ^ ... ^ aj = sum (i-1) ^ sum (j).

Next we can follow the query interval with the MO team algorithm to traverse through, while recording the current interval a prefix and the number of occurrences;

According to sum (i-1) ^ sum (j) = k, k ^ sum (i-1) = SUM (j) | | K ^ sum (j) = SUM (i-1), the number of qualifying (I,J) is introduced by the prefix and number of times that match the condition.

TIP:

Listening to the lesson of the great God, this problem has two pits:

1, the data of the answer data range is burst int;

2, although the 1e6 K, but the result of the XOR is greater than 1e6;

AC Code:

1#include <bits/stdc++.h>2 #defineLL Long Long int3 using namespacestd;4 Const intMAXN =1<< -;5 6 structnode7 {8 intL, R, id;//interval and query number9}Q[MAXN];//Record query data (offline)Ten One intPOS[MAXN];//record chunking ALL ANS[MAXN];//Record Answers -LL FLAG[MAXN];//maintain prefix XOR and occurrences - intA[MAXN];//Original Data the intL=1R//the left and right nodes of the current interval -LL Res;//stores the value of the current interval - intN, M, K; - BOOLCMP (Node A, Node B)//Sort + { - if(POS[A.L]==POS[B.L])returnA.R < B.R;//only the left node is the right node in the same block . + returnPOS[A.L] < POS[B.L];//Otherwise, the left node is divided into blocks. A } at voidAddintx) - { -res+=flag[a[x]^K]; -flag[a[x]]++; - } - voidDelintx) in { -flag[a[x]]--; tores-=flag[a[x]^K]; + } - intMain () the { *scanf"%d%d%d", &n, &m, &K); $ intSZ =sqrt (N);Panax Notoginseng for(inti =1; I <= N; i++) {//read in Data -scanf"%d", &a[i]); theA[i] = a[i]^a[i-1];//computes the prefix XOR or +Pos[i] = I/sz;// chunking A } the for(inti =1; I <= M; i++) {//read-in Query +scanf"%d%d", &Q[I].L, &Q[I].R); -Q[i].id =i; $ } $Sort (q+1, q+1+M, CMP); -flag[0] =1; - for(inti =1; I <= M; i++){ the while(L < Q[I].L) {//The current left node is smaller than the query node. -Del (L-1);Wuyil++; the } - while(L > Q[I].L) {//The current left node is larger than the left node of the query. Wul--; -Add (l1); About } $ while(R < Q[I].R) {//The current right node is smaller than the query right node. -r++; - Add (R); - } A while(R > Q[I].R) {//Current right node is larger than query right node + del (R); ther--; - } $Ans[q[i].id] =Res; the } the for(inti =1; I <= M; i++){ theprintf"%lld\n", Ans[i]); the } -}

View Code

Codeforces Round #340 (Div. 2) E. XOR and favorite number "Mo Team algorithm + xor and prefix and ingenious"