E. XOR and Favorite Number（莫队算法）

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.

Sample test(s)

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：有n个数和m次查询，每次查询区间[l, r]问满足ai ^ ai+1 ^ ... ^ aj == k的(i, j) (l <= i <= j <= r)有多少对。

利用前缀和的思想在传入值num的时候就计算异或（num[i]=num[i]^num[i-1]）;类似sum[i]=sum[i-1]+a[i]，对于add与del函数

由于，pre[]<--->flag[] 中保存的是前缀和，所以num[x]^k的位置就是满足能亦或等到 K 的所有对数所在的位置，所以有flag[num[x]^k]。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <map>
#include <cmath>
#include <vector>
#include <string>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
const int MAX1=0x7f7f7f7f;
const int MAXN=5e6+10;
const int MAX=1e5+10;
typedef long long LL;

struct node{
int l,r,id;
}Q[MAXN];
LL num[MAXN]; 
LL flag[MAXN]; //前缀和出现的次数
LL pos[MAXN];  //块的划分
LL Ans;
LL ans[MAXN];
int L,R;
LL n,m,k;
bool cmp(node a,node b) //分块排序
{
    if(pos[a.l]==pos[b.l])
        return a.r<b.r;
    return pos[a.l]<pos[b.l];
}
void add(int  x)
{
    Ans+=flag[num[x]^k]; 
    flag[num[x]]++;
}
void del(int  x)
{
    flag[num[x]]--;  //计算时可能是其本身，需要提前去掉
    Ans-=flag[num[x]^k];
    
}
int sz;
int main()
{
    ios::sync_with_stdio(false);
     cin>>n>>m>>k;
     memset(pos,0,sizeof(pos));
     memset(num,0,sizeof(num));
     memset(flag,0,sizeof(flag));
    Ans=0;
    L=1,R=0;
    sz=sqrt(n);//分块的大小
    flag[0]=1; //就算不加入值也还是有值的
    for(int i=1;i<=n;i++)
    {
        cin>>num[i];
        pos[i]=i/sz;
        num[i]=num[i]^num[i-1];
    }

    for(int i=1;i<=m;i++)
    {
        cin>>Q[i].l>>Q[i].r;
        Q[i].id=i;
    }
    sort(Q+1,Q+1+m,cmp);

    for(int i=1;i<=m;i++)
    {
        while(L<Q[i].l)//L在要查询的区间左边,则把L右移 而ans包含的是【L R】的值
        {//所以应该先减去以前的 L 位置的数形成的组合数重新构造
            del(L-1);//处理的是pre[L-1]->pre[R] 所以会L-1
            L++;
        }
        while(L>Q[i].l)
        {
            L--;
            add(L-1);
        }
        while(R<Q[i].r)
        {
            R++;
            add(R);
        }
        while(R>Q[i].r)
        {
            del(R);
            R--;
        }

        ans[Q[i].id]=Ans;
    }
    for(int i=1;i<=m;i++)
        cout<<ans[i]<<endl;

return 0;
}

E. XOR and Favorite Number（莫队算法）

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