Codeforces Round #340 (Div. 2) 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

5 3 1
1 1 1 1 1
1 5
2 4
1 3

output

9
4

4

题意：给你n个数和m个询问以及k，每个询问给一个区间[l,r],问区间内有多少对(i,j),使得a[i]^a[i+1]^a[i+2]^...^a[j]=k.

思路：我们可以求出前缀异或和a[i]，那么题目就变成问区间[l,r]内，有多少对(i,j)满足a[i-1]^a[j]=k,我们可以用cnt[i]记录异或值为i的个数，然后就能用莫队算法了。

#include<stdio.h>
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef __int64 ll;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 100050
ll a[maxn],sum[maxn],unit;
struct node{
    ll l,r,idx;
}b[maxn];
bool cmp(node a,node b){
    if(a.l/unit == b.l/unit){
        return a.r < b.r;
    }
    return a.l/unit < b.l/unit;
}

ll cnt[1050000];
ll ans[maxn];

int main()
{
    ll n,m,k;
    int i,j;
    while(scanf("%I64d%I64d%I64d",&n,&m,&k)!=EOF)
    {
        a[0]=0;
        for(i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
            a[i]=a[i-1]^a[i];
        }
        unit=(ll)sqrt(n);
        for(i=1;i<=m;i++){
            scanf("%I64d%I64d",&b[i].l,&b[i].r);
            b[i].idx=i;
        }
        sort(b+1,b+1+m,cmp);
        ll l=1,r=0;
        ll num=0;
        memset(cnt,0,sizeof(cnt));
        cnt[0]=1;  //这里要注意，一开始cnt[0]=0
        for(i=1;i<=m;i++){    //对于询问区间[l,r],相当于在维护【a[l-1],a[r]】出现的次数。
            while(r<b[i].r){  //每个while语句里的前后顺序要注意
                 r++;
                 num+=cnt[k^a[r] ];
                 cnt[a[r] ]++;
            }
            while(r>b[i].r){
                cnt[a[r] ]--;
                num-=cnt[k^a[r] ];
                r--;
            }
            while(l<b[i].l){
                cnt[a[l-1] ]--;
                num-=cnt[k^a[l-1] ];
                l++;
            }
            while(l>b[i].l){
                l--;
                num+=cnt[k^a[l-1] ];
                cnt[a[l-1] ]++;

            }
            ans[b[i].idx ]=num;
        }
        for(i=1;i<=m;i++){
            printf("%I64d\n",ans[i]);
        }
    }
    return 0;
}