Codeforces 617E XOR and Favorite Number 【莫队】

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

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;
}