Submission #78500

#	Time	Username	Problem	Language	Result	Execution time	Memory
78500		youngyojun	XOR (IZhO12_xor)	C++11	100 / 100	287 ms	73580 KiB

xor

#include <bits/stdc++.h>
#define rf(x) (x)=0;while(*p<48)p++;while(47<*p)(x)=((x)<<3)+((x)<<1)+(*p++&15);
//#define rf(x) (x)=0;while(*p<48)im=*p=='-';while(47<*p)(x)=((x)<<3)+((x)<<1)+(*p++&15);if(im)(x)=-(x);
#define pb push_back
#define eb emplace_back
#define sz(V) ((int)(V).size())
#define allv(V) ((V).begin()),((V).end())
#define befv(V) ((V)[(sz(V)-2)])
#define sorv(V) sort(allv(V))
#define revv(V) reverse(allv(V))
#define univ(V) (V).erase(unique(allv(V)),(V).end())
#define clv(V) (V).clear()
#define upmin(a,b) (a)=min((a),(b))
#define upmax(a,b) (a)=max((a),(b))
#define rb(x) ((x)&(-(x)))
#define cb(x) (x)=(!(x))
#define INF (0x3f3f3f3f)
#define INFLL (0x3f3f3f3f3f3f3f3fll)
#define INFST (0x7f7f7f7f)
#define INFLLST (0x7f7f7f7f7f7f7f7fll)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, ll> pil;
typedef pair<ll, int> pli;
typedef pair<ld, ld> pdd;
typedef complex<ld> base;
const ld EPS = (ld)1e-7;
const ld PI = acos(0) * 2;
bool isZero(const ld& x) { return abs(x) <= EPS; }
int sign(const ld& x) { return isZero(x) ? 0 : (0 < x ? 1 : -1); }
ll gcd(ll a, ll b) { for(;b;a%=b,swap(a,b)){} return abs(a); }
pll operator + (const pll& a, const pll& b) { return pll(a.first+b.first, a.second+b.second); }
pll operator - (const pll& a, const pll& b) { return pll(a.first-b.first, a.second-b.second); }
pll operator * (const pll& a, const ll& b) { return pll(a.first*b, a.second*b); }
ll operator * (const pll& a, const pll& b) { return a.first*b.second - b.first*a.second; }
ll ccw(const pll& a, const pll& b, const pll& c) { return a*b + b*c + c*a; }
void fg(vector<int> G[], int a, int b) { G[a].pb(b); G[b].pb(a); }
void fg(vector<pii> G[], int a, int b, int c) { G[a].pb({b, c}); G[b].pb({a, c}); }
int operator * (pii a, pii b) { return a.first*b.second - a.second*b.first; }
int ccw(pii a, pii b, pii c) { return a*b + b*c + c*a; }

const int MAXN = 300055;

struct SEG {
	SEG() : l(0), r(0), dt(INF) {}
	SEG *l, *r;
	int dt;

	void upd(int s, int e, int x, int i) {
		upmin(dt, i);
		if(s == e) return;
		const int m = (ll(s)+e) >> 1;
		if(x <= m) {
			if(!l) l = new SEG();
			l -> upd(s, m, x, i);
		} else {
			if(!r) r = new SEG();
			r -> upd(m+1, e, x, i);
		}
	}

	int get(int s, int e, int x, int y, int key) {
		if(s == e) return y <= (s^x) ? dt : INF;
		const int m = (ll(s)+e) >> 1;
		if(y & key) {
			if(x & key) return l ? (l -> get(s, m, x, y, key >> 1)) : INF;
			return r ? (r -> get(m+1, e, x, y, key >> 1)) : INF;
		} else {
			int ret = INF;
			if(x & key) {
				if(l) ret = l -> dt;
				if(r) upmin(ret, r -> get(m+1, e, x, y, key >> 1));
			} else {
				if(r) ret = r -> dt;
				if(l) upmin(ret, l -> get(s, m, x, y, key >> 1));
			}
			return ret;
		}
	}
} *rt = new SEG();

int A[MAXN], B[MAXN];

int N, X, AnsI = -INF, AnsL;

int main() {
	ios::sync_with_stdio(false);

	cin >> N >> X;
	for(int i = 1; i <= N; i++) {
		cin >> A[i];
		B[i] = B[i-1] ^ A[i];
	}

	for(int i = 1, l; i <= N; i++) {
		rt -> upd(0, 2147483647, B[i-1], i-1);
		int j = rt -> get(0, 2147483647, B[i], X, 1 << 30);
		if(i <= j) continue;
		l = i - j;
		if(AnsL < l) {
			AnsL = l;
			AnsI = j+1;
		}
	}

	if(AnsI <= 0) puts("-1");
	else printf("%d %d\n", AnsI, AnsL);
	return 0;
}

• Subtask 1: 100 / 100