Codeforces Round #340 (Div. 2) E. XOR and Favorite Number 莫队算法-优快云博客

本文介绍了一个编程问题的解决方法，该问题是关于在一个整数序列中进行区间异或操作的查询。通过对莫队算法的应用，文章提供了一种有效的解决方案，并详细解释了实现过程。

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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

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

　　m个询问，每次询问你l,r之间有多少对子区间的异或和等于k

题解：

　　怼一波莫队

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+10, M = 5e6+10, inf = 1e9+7, mod = 1e9+7;

int  H[M];
int sum[N],n,m,k,a[N],pos[N];
long long  an[N],ans;
struct ss{int l,r,id;}q[N];

int cmp(ss s1,ss s2) {if(pos[s1.l]==pos[s2.l]) return s1.r<s2.r;else return s1.l<s2.l;}
int cmpl(ss s1,ss s2) {return s1.id<s2.id;}
void update(int p,int add) {
    if(add==-1) {
        H[sum[p]]--;
        ans -= H[sum[p]^k];
    }
    else {
        ans += H[sum[p]^k];
        H[sum[p]]++;
    }
}
int main() {
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    H[0] = 1;
    for(int i=1;i<=n;i++) sum[i] = sum[i-1] ^ a[i];
    for(int i=1;i<=m;i++) {
        scanf("%d%d",&q[i].l,&q[i].r);
        q[i].id = i;
    }
    int block = (int)sqrt((double) n + 0.5);
    for(int i=1;i<=n;i++) pos[i] = (i-1)/block  + 1;
    sort(q+1,q+m+1,cmp);
    for(int i=1,l=1,r=0;i<=m;i++) {
        for(;r<q[i].r;r++) update(r+1,1);
        for(;l>q[i].l;l--) update(l-2,1); 
        for(;r>q[i].r;r--) update(r,-1);
        for(;l<q[i].l;l++) update(l-1,-1);
        an[q[i].id] = ans;
    }
    for(int i=1;i<=m;i++) printf("%I64d\n",an[i]);
    return 0;
}