Beautiful Subarrays 　　

题解：

把数字转化成2进制之后，用字典树去维护。

 

想到字典树之后就应该是一道很容易写的题目了。

 

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e6 + 100;
int n, k;
int a[N];
int s[30], ss[30];
int tree[1<<24][2];
int cnt[1<<24];
int tot = 2;
void add(){
    int rt = 1;
    for(int i = 0; i < 30; ++i){
        int id = s[i];
        if(!tree[rt][id]) tree[rt][id] = ++tot;
        ++cnt[tree[rt][id]];
        rt = tree[rt][id];
    }
}
void add(int x){
    for(int i = 29; i >= 0; --i){
        s[i] =  + (x & 1);
        x >>= 1;
    }
    add();
}
LL ans = 0;
void Find(){
    int rt = 1;
    for(int i = 0; i < 30; ++i){
        if(!rt) return ;
        if(s[i] == 0 && ss[i] == 0){
//            cout << " *** " << endl;
            ans += cnt[tree[rt][1]];
            rt = tree[rt][0];
        }
        else if(s[i] == 0 && ss[i] == 1){
            rt = tree[rt][1];
        }
        else if(s[i] == 1 && ss[i] == 0){
//            cout << "??" << endl;
            ans += cnt[tree[rt][0]];
            rt = tree[rt][1];
        }
        else if(s[i] == 1 && ss[i] == 1){
            rt = tree[rt][0];
        }
//        cout << i << " " << ss[i] << " " <<ans << endl;
    }
    ans += cnt[rt];
}

int main(){
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
    for(int i = 29; i >= 0; --i){
        ss[i] =  + (k & 1);
        k >>= 1;
    }
    add(0);
    for(int i = 1; i <= n; ++i){
        a[i] ^= a[i-1];
        add(a[i]);
        Find();
    }
    cout << ans << endl;
    return 0;
}

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