Educational Codeforces Round 12 E. Beautiful Subarrays(xor trie)

题意:

$N\le 10^6个点的数,xor(l, r)=A_l\oplus A_{l+1}\oplus\cdots\oplus A_r$
$求xor(l, r)\ge k的(l, r)对数$

分析:

$xor trie贪心$
$我们发现这个是个连续异或和,经典的技巧$
$设prefix[i]:=A_1\oplus A_2\oplus\cdots\oplus A_i$
$那么显然我们可以得到xor(l, r)=prefix[r]\oplus prefix[l-1]$
$所以就枚举每个r,不断的把prefix[i]从高位到低位插入到trie里$
$然后贪心的去找大于k的,因为插入的时候把这个prefix[i]的所有节点的cnt都增加了$
$最后再把=k的加上就可以不重不漏的算完了$
$时间复杂度O(nb)$

代码:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
//
// Created by TaoSama on 2016-04-21
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
struct Trie {
static const int M = 32 * 1e6 + 10, S = 2;
int root, sz;
int nxt[M][S], cnt[M];
int newNode() {
cnt[sz] = 0;
memset(nxt[sz], -1, sizeof nxt[sz]);
return sz++;
}
void init() {
sz = 0;
root = newNode();
}
void insert(int x) {
int u = root;
for(int i = 31; ~i; --i) {
int c = x >> i & 1, &v = nxt[u][c];
if(v == -1) v = newNode();
++cnt[v];
u = v;
}
}
int query(int x, int k) {
int u = root, ret = 0;
for(int i = 31; ~i; --i) {
int c = x >> i & 1;
if(k >> i & 1) u = nxt[u][c ^ 1];
else {
ret += cnt[nxt[u][c ^ 1]]; //>
u = nxt[u][c];
}
if(u == -1) return ret;
}
return ret + cnt[u]; //=
}
} trie;
int n, k;
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d%d", &n, &k) == 2) {
trie.init();
trie.insert(0);
int sum = 0;
long long ans = 0;
for(int i = 1; i <= n; ++i) {
int x; scanf("%d", &x);
sum ^= x;
ans += trie.query(sum, k);
trie.insert(sum);
}
printf("%I64d\n", ans);
}
return 0;
}


1. 除非注明,本博文即为原创,转载请注明链接地址
2. 本博文只代表博主当时的观点或结论,请不要恶意攻击
3. 如果本文帮到了您,不妨点一下 下面分享到 按钮,让更多的人看到