AT4142 [ARC098B] Xor Sum 2

(large{题目链接})
(\)
(Large extbf{Solution: } large{首先，异或有一个性质： ext{x} - ext{y} leq ext{x} oplus ext{y} leq ext{x} + ext{y}。\观察题目，容易发现，若一个区间符合要求，那么 ext{a}_l到 ext{a}_y的二进制每一位上最多只有一个1。\然后设f[l]为l所能拓展到的最右点，那么显然l + 1也符合条件，因为l+1中1的个数只可以比l少，所以l满足单增，可以用尺取法来维护。})

(Large extbf{Code: })

#include <bits/stdc++.h>
#define gc() getchar() 
#define LL long long
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define _rep(i, a, b) for (int i = (a); i >= (b); --i)
using namespace std;
const int N = 2e5 + 5;
int n, a[N];

inline int read() {
	int x = 0;
	char ch = gc();
	while (!isdigit(ch)) ch = gc();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = gc();
	return x; 
}

int main() {
	ios::sync_with_stdio(false);
	n = read();
	rep(i, 1, n) a[i] = read();
	LL ans = 0;
	for (int i = 1, l = 1, r = 1, cur = 0; i <= n; ++i) {
		for (; !(cur & a[r]) && r <= n;) cur |= a[r++], ans += r - l;
		cur ^= a[l++];
	}
	cout << ans << endl;
	return 0;
}