Codeforces 617E XOR and Favorite Number 【莫队】

最新推荐文章于 2021-10-14 13:38:53 发布

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最新推荐文章于 2021-10-14 13:38:53 发布

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分类专栏： codeforces 莫队

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本文介绍了一种解决区间异或查询问题的有效算法。通过预处理前缀异或和及使用莫队算法优化查询过程，实现了快速计算指定区间内元素异或结果等于特定值的子数组对数量。

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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.

   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

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。

思路：离线做。首先预处理前缀异或和sum[]，那么ai ^ ... ^ aj == sum[i-1] ^ sum[j]。

这样对一次查询[l, r]的处理，可以从左到右扫一次，统计k ^ sum[i]出现的次数(l <= i <= r)。

假设已经处理到[L, R]，对下一次的[l, r]处理——

若L < l，显然多余，需要去掉[L, l-1]的部分，若L > l需要加上[l, L-1]的部分，反之不需要处理。

若R > r，..................[r+1, R]......，若R < r........[R+1, r]......，..............。

用莫队做即可。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (100000+10)
#define MAXM (200000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
using namespace std;
struct Node{
    int l, r, id; LL ans;
};
Node num[MAXN];
bool cmp1(Node a, Node b)
{
    if(a.l / 400 != b.l / 400)
        return a.l / 400 < b.l / 400;
    else
        return a.r < b.r;
}
bool cmp2(Node a, Node b){
    return a.id < b.id;
}
LL cnt[20000000+10];
int sum[MAXN];
int main()
{
    int n, m, k;
    Ri(n); Ri(m); Ri(k); int a;
    for(int i = 1; i <= n; i++)
        Ri(a), sum[i] = sum[i-1] ^ a;
    for(int i = 1; i <= m; i++)
        Ri(num[i].l), Ri(num[i].r), num[i].id = i, num[i].l--;
    sort(num+1, num+m+1, cmp1);
    LL s = 0; int L = num[1].l, R = num[1].r;
    for(int i = L; i <= R; i++)
    {
        s += cnt[k ^ sum[i]];
        cnt[sum[i]]++;
    }
    num[1].ans = s;
    for(int i = 2; i <= m; i++)
    {
        while(L > num[i].l)
        {
            L--;
            s += cnt[k ^ sum[L]];
            cnt[sum[L]]++;
        }
        while(L < num[i].l)
        {
            //L++;
            cnt[sum[L]]--;
            s -= cnt[k ^ sum[L]];
            L++;
        }
        while(R < num[i].r)
        {
            R++;
            s += cnt[k ^ sum[R]];
            cnt[sum[R]]++;
        }
        while(R > num[i].r)
        {
            //R--;
            cnt[sum[R]]--;
            s -= cnt[k ^ sum[R]];
            R--;
        }
        num[i].ans = s;
    }
    sort(num+1, num+m+1, cmp2);
    for(int i = 1; i <= m; i++) Pl(num[i].ans);
    return 0;
}