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 thatl ≤ i ≤ j ≤ r and the xor of the numbers ai, ai + 1, ..., aj is equal to k.
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.
Print m lines, answer the queries in the order they appear in the input.
6 2 3
1 2 1 1 0 3
1 6
3 5
7
0
5 3 1
1 1 1 1 1
1 5
2 4
1 3
9
4
4
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次询问,每一询问会有一个L和R,表示所询问的区间,
问在这个区间中有多少个连续的子区间的亦或和为k
假设我们现在有一个前缀异或和数组sum[],现在我们要求区间[L,R]的异或的值,
用sum数组表示就是sum[L-1]^sum[R]==K,或者说是K^sum[R]==sum[L-1]
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 2e6 + 10; 5 int n, m, k, L, R, sz, a[maxn]; 6 LL sum[maxn], ans, ANS[maxn]; 7 struct node { 8 int l, r, id; 9 node() {} 10 node(int l, int r, int id): l(l), r(r), id(id) {} 11 bool operator <(const node & a)const { 12 if (l / sz == a.l / sz) return r < a.r; 13 return l < a.l; 14 } 15 } qu[maxn]; 16 void add(int x) { 17 ans += sum[a[x] ^ k]; 18 sum[a[x]]++; 19 } 20 void del(int x) { 21 sum[a[x]]--; 22 ans -= sum[a[x] ^ k]; 23 } 24 int main() { 25 scanf("%d%d%d", &n, &m, &k); 26 for (int i = 1 ; i <= n ; i++) { 27 scanf("%d", &a[i]); 28 a[i] ^= a[i - 1]; 29 } 30 for (int i = 1 ; i <= m ; i++) { 31 scanf("%d%d", &qu[i].l, &qu[i].r); 32 qu[i].l--; 33 qu[i].id = i; 34 } 35 sz = (int)sqrt(n); 36 sort(qu + 1, qu + m + 1); 37 L = 1, R = 0; 38 for (int i = 1 ; i <= m ; i++) { 39 while(L > qu[i].l) add(--L); 40 while(R < qu[i].r) add(++R); 41 while(L < qu[i].l) del(L++); 42 while(R > qu[i].r) del(R--); 43 ANS[qu[i].id] = ans; 44 } 45 for (int i = 1 ; i <= m ; i++) 46 printf("%lld ", ANS[i]); 47 return 0; 48 }