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  • XOR and Favorite Number (莫对算法)

    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 thatl ≤ 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 nm 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.

    Examples
    input
    Copy
    6 2 3
    1 2 1 1 0 3
    1 6
    3 5
    output
    Copy
    7
    0
    input
    Copy
    5 3 1
    1 1 1 1 1
    1 5
    2 4
    1 3
    output
    Copy
    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 }
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  • 原文地址:https://www.cnblogs.com/qldabiaoge/p/9360505.html
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