제출 #30922

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
30922		sean617	Gap (APIO16_gap)	C++98	30 / 100	79 ms	11600 KiB

gap

#include "gap.h"
#define SZ 100005
#include <iostream>
#include <algorithm>
#include <vector>
typedef long long ll;

using namespace std;
ll n, x = -1, num = 1e18 + 5, a[SZ];
vector<ll> b;

void f(ll p, ll q) {
	ll mn1, mx1, mn2, mx2;
	mn1 = mx1 = mn2 = mx2 = -1;
	if (p > q || b.size() == n || q - p <= x) return;
	if (q - p <= num) {
		MinMax(p, q, &mn1, &mx1);
		if (mn1 == -1) return;
		b.push_back(mn1);
		if (mn1 != mx1) b.push_back(mx1);
		f(mn1 + 1, mx1 - 1);
	} else {
		ll md = (p + q) / 2;
		MinMax(p, md, &mn1, &mx1);
		if (p < q) MinMax(md + 1, q, &mn2, &mx2);
		if (mn1 == -1 && mx1 == -1 && mn2 == -1 && mx2 == -1) x = max(x, (q + 1) - (p - 1));
		if (mn1 != -1) {
			b.push_back(mn1);
			if (mn1 != mx1) b.push_back(mx1);
			f(mn1 + 1, mx1 - 1);
		}
		if (p < q && mn2 != -1) {
			b.push_back(mn2);
			if (mn2 != mx2) b.push_back(mx2);
			f(mn2 + 1, mx2 - 1);
		}
	}
}

long long findGap(int T, int N)
{
	ll st = 0, en = 1e18, i = 0, j = N - 1, mn, mx, ans;
	n = N;
	if (T == 1) {
		while (i <= j) {
			MinMax(st, en, &mn, &mx);
			a[i] = mn;
			a[j] = mx;
			st = mn + 1;
			en = mx - 1;
			i++;
			j--;
		}
		ans = a[1] - a[0];
		for (i = 1; i < N - 1; i++) {
			ans = max(ans, a[i + 1] - a[i]);
		}
	} else {
		MinMax(st, en, &mn, &mx);
		b.push_back(mn);
		b.push_back(mx);
		x = (mx - mn) / (N - 1);
		if ((mx - mn) % (N - 1) > 0) x++;
		f(mn + 1, mx - 1);
		sort(b.begin(), b.end());
		ans = x;
		for (i = 0; i < b.size() - 1; i++) {
			ans = max(ans, b[i + 1] - b[i]);
		}
	}
	return ans;
}

컴파일 시 표준 에러 (stderr) 메시지

gap.cpp: In function 'void f(ll, ll)':
gap.cpp:15:24: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
  if (p > q || b.size() == n || q - p <= x) return;
                        ^
gap.cpp: In function 'long long int findGap(int, int)':
gap.cpp:67:17: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
   for (i = 0; i < b.size() - 1; i++) {
                 ^

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