제출 #1055808

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1055808		thieunguyenhuy	Gap (APIO16_gap)	C++17	0 / 100	45 ms	1248 KiB

gap

#ifndef hwe
	#include "gap.h"
#endif

#include <bits/stdc++.h>
using namespace std;

#define popcount(n) (__builtin_popcountll((n)))
#define clz(n) (__builtin_clzll((n)))
#define ctz(n) (__builtin_ctzll((n)))
#define lg(n) (63 - __builtin_clzll((n)))
#define BIT(n, i) (((n) >> (i)) & 1ll)
#define MASK(i) (1ll << (i))
#define FLIP(n, i) ((n) ^ (1ll << (i)))
#define ON(n, i) ((n) | MASK(i))
#define OFF(n, i) ((n) & ~MASK(i))

#define Int __int128
#define fi first
#define se second

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef pair<long long, int> pli;
typedef pair<int, long long> pil;
typedef vector<pair<int, int>> vii;
typedef vector<pair<long long, long long>> vll;
typedef vector<pair<long long, int>> vli;
typedef vector<pair<int, long long>> vil;

template <class T1, class T2>
bool maximize(T1 &x, T2 y) {
    if (x < y) {
        x = y;
        return true;
    }
    return false;
}
template <class T1, class T2>
bool minimize(T1 &x, T2 y) {
    if (x > y) {
        x = y;
        return true;
    }
    return false;
}

template <class T>
void remove_duplicate(vector<T> &ve) {
    sort (ve.begin(), ve.end());
    ve.resize(unique(ve.begin(), ve.end()) - ve.begin());
}

mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
template <class T> T random(T l, T r) {
    return uniform_int_distribution<T>(l, r)(rng);
}
template <class T> T random(T r) {
    return rng() % r;
}

const int N = 1e6 + 5;
const int MOD = 1e9 + 7;
const int inf = 1e9;
const ll INF = 1e18;

#ifdef hwe
void MinMax(ll s, ll t, ll *mn, ll *mx) {}
#endif

int findGap(int t, int n) {
	ll mi = -1, ma = -1;
	MinMax(0, INF, &mi, &ma);

	ll L = ma - mi, block_size = (L + n - 2) / (n - 1);

	ll last_max = -1, ans = -INF;
	for (ll i = mi; i <= ma; i += block_size) {
		ll mn = -1, mx = -1;
		MinMax(i, min(i + block_size - 1, ma), &mn, &mx);
		if (mn != -1 && last_max != -1) maximize(ans, mi - last_max);
		if (mx != -1) last_max = mx;
	}

	return ans;
}

#ifdef hwe
signed main() {



    cerr << '\n'; return 0;
}
#endif

• Subtask 1: 0 / 30
• Subtask 2: 0 / 70