题意：
- 给一个长度为n的数组a，问有多少个连续的子数组的异或和大于等于k
思路：
- 01字典树
- 建立前缀异或和数组，题目变为有多少个$a_i ˆ a_j(1 \leq i < j \leq n)$
- 类似（几乎一模一样）的题目
-
一次插入一个左端点，然后枚举右端点的情况
-
R ^ L $\leq$ k，从高位往低位
-
$k_i = 1$，必须$R_i$ ^ $L_i = 1$，走 $R_i $ ^ $ 1$；$R_i$ ^ $L_i = 0$会造成比 k 小，忽略
- $k_i = 0$，若$R_i$ ^ $L_i = 1$，加上$R_i $ ^ $ 1 $这个方向，因为这个方向必然会比 k 小；若$R_i$ ^ $L_i = 0$，走$R_i $这个方向
#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<" "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<" "<<#d<<" = "<<d<<" "<<#e<<" = "<<e<<endl;
#define endl "\n"
#define fi first
#define se second
//#define int long long
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)
const int N = 1e6+10;
int son[32*N][2],idx;
int cnt[32*N];
int a[N];
int n,k;
void insert(int x)
{
int p = 0;
for (int i = 30; i >= 0; i -- )
{
int &s = son[p][x >> i & 1];
if (!s) s = ++ idx;
p = s;
cnt[p] ++;
}
}
int get(int x,int R)
{
int p = 0;
int res = 0;
for(int i = 30;i >= 0;i -- )
{
if(x >> i & 1)
{
p = son[p][(R>>i&1)^1];
}else{
res += cnt[son[p][(R>>i&1)^1]];
p = son[p][R>>i&1];
}
if(p == 0)return res;
}
return res + cnt[p];
}
void solve()
{
cin >> n >> k;
for(int i = 1;i <= n;i ++)
{
cin >> a[i];
a[i] ^= a[i-1];
}
long long ans = 0;
insert(0);
for(int i = 1;i <= n;i ++)
{
ans += get(k,a[i]);
insert(a[i]);
}
cout << ans << endl;
}
signed main()
{
/*
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
*/
int T = 1;//cin >> T;
while(T--){
//puts(solve()?"YES":"NO");
solve();
}
return 0;
}
/*
*/
标签：Beautiful,cnt,01,return,int,Subarrays,son,res From： https://www.cnblogs.com/cfddfc/p/17044702.html