제출 #1158142

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1158142		yonatanl	XOR Sum (info1cup17_xorsum)	C++20	0 / 100	1697 ms	25688 KiB

xorsum

#include <iostream>
#include <vector>
#include <algorithm>

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")

#define rep(i, s, e) for (ll i = s; i < e; i++)

using namespace std;
using ll = long long;
using vll = vector<ll>;
using vb = vector<bool>;

void solve() {
	ll n;
	cin >> n;
	vll arr(n);
	rep(i, 0, n) cin >> arr[i];
	ll ans = 0;
	rep(b, 0, 28) {
		// Look at the first b bits
		vll cur0; // Keep all the numbers that have 0 in the b bit with only their first b-1 bits
		vll cur1; // Keep all the numbers that have 1 in the b bit with only their first b-1 bits
		ll A = ((ll)1 << (b)) - 1;
		rep(i, 0, n) {
			if ((arr[i] >> b) & (ll)1) cur1.push_back(arr[i] & A);
			else cur0.push_back(arr[i] & A);
		}
		sort(cur0.begin(), cur0.end());
		sort(cur1.begin(), cur1.end());
		ll res = 0;
		ll carry = (ll)1 << b;
		rep(i, 0, n) {
			ll curVal = arr[i] & A; // The first b-1 bits of arr[i]
			ll num0, num1;
			if (((2 * arr[i]) >> b) & (ll)1) res++;
			if (((arr[i] >> b) & (ll)1) == 0) {
				// The b bit is 0
				num0 = 0; // The number of numbers that have 0 in the b bit and give carry
				num1 = 0; // The number of numbers that have 1 in the b bit and don't give carry
				num0 = cur0.end() - lower_bound(cur0.begin(), cur0.end(), carry - curVal);
				num1 = lower_bound(cur1.begin(), cur1.end(), carry - curVal) - cur1.begin();
			}
			else {
				// The b bit is 1
				num0 = 0; // The number of numbers that have 0 in the b bit and don't give carry
				num1 = 0; // The number of numbers that have 1 in the b bit and give carry
				num0 = lower_bound(cur0.begin(), cur0.end(), carry - curVal) - cur0.begin();
				num1 = cur1.end() - lower_bound(cur1.begin(), cur1.end(), carry - curVal);
			}
			res += (num0 + num1);
		}
		res /= 2;
		ans += (res % 2) * carry;
	}
	cout << ans << endl;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	solve();
}

• Subtask 1: 0 / 7
• Subtask 2: 0 / 11
• Subtask 3: 0 / 21
• Subtask 4: 0 / 38
• Subtask 5: 0 / 23