CodeForces - 617E XOR and Favorite Number （莫队+前缀和）

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.

Examples

Input

6 2 31 2 1 1 0 31 63 5

Output

Input

5 3 11 1 1 1 11 52 41 3

Output

944

题意：询问区间内异或和刚好为k的字段个数。思路：莫队+前缀和。这个前缀和比较套路，用的是前缀异或和。字段【l,r】的异或和就是pre[r]^pre[l-1],这种情况下我们在莫队的过程中记录l，r的pre[i]出现的次数，就可以完成更新了。

注意当L<q[i].l时，要先让记录per[L]出现次数的个数减一。

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int maxm = 1000086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int num[maxm*2],pre[maxn],a[maxn];struct node{int l,r;int id;
}q[maxn];
ll ans[maxn];
ll anss;
int block;bool cmp(node a,node b){if(a.l/block!=b.l/block){return a.l<b.l;}return a.r<b.r;
}int main()
{int n,m,k;scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=n;i++){scanf("%d",&a[i]);pre[i]=pre[i-1]^a[i];}block=sqrt(n);for(int i=1;i<=m;i++){scanf("%d%d",&q[i].l,&q[i].r);q[i].id=i;}sort(q+1,q+1+m,cmp);int L=1,R=0;anss=0;num[0]=1;for(int i=1;i<=m;i++){while(L<q[i].l){int t=k^pre[L-1];num[pre[L-1]]--;///注意语句顺序anss-=num[t];L++;}while(R>q[i].r){int t=k^pre[R];num[pre[R]]--;anss-=num[t];R--;}while(L>q[i].l){L--;int t=k^pre[L-1];anss+=num[t];num[pre[L-1]]++;}while(R<q[i].r){R++;int t=k^pre[R];anss+=num[t];num[pre[R]]++;}ans[q[i].id]=anss;}for(int i=1;i<=m;i++){printf("%lld\n",ans[i]);}return 0;
}

View Code

转载于:https://www.cnblogs.com/ZGQblogs/p/10863851.html

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