제출 #1076594

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1076594		Pannda	Gap (APIO16_gap)	C++17	100 / 100	41 ms	3616 KiB

gap

#include "gap.h"

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

long long findGap(int type, int n) {
    constexpr long long C = 1e18;
    if (type == 1) {
        vector<long long> a(n);
        int idx_l = 0, idx_r = n - 1;
        long long l = 0, r = C;
        for (int t = 0; t < (n + 1) / 2; t++) {
            MinMax(l, r, &a[idx_l], &a[idx_r]);
            l = a[idx_l] + 1;
            r = a[idx_r] - 1;
            idx_l++;
            idx_r--;
        }
        long long ans = 0;
        for (int i = 0; i + 1 < n; i++) {
            ans = max(ans, a[i + 1] - a[i]);
        }
        return ans;
    } else {
        long long x;
        long long last;
        MinMax(0, C, &x, &last);
        long long ans = (last - x) / (n - 1);

        while (last - x > ans) {
            long long mn, mx;
            MinMax(x + 1, x + ans, &mn, &mx);
            if (mn == -1) {
                break;
            } else {
                x = mx;
            }
        }

        while (last - x > ans) {
            // with x, find some d such that [x, x + d] is only x, and [x + d + 1, x + d + 1 + d] is none empty and is (mn, mx)
            // update ans with mn - x, and set new x to mx
            // jump x until [x + ans] is only x
            while (last - x > ans) {
                // assert( [x, x + ans] is only x)
                long long s = x + ans + 1;
                assert(s <= C);
                long long mn, mx;
                MinMax(s, s + ans, &mn, &mx);
                if (mn == -1) {
                    ans = ans + ans + 1;
                } else {
                    ans = mn - x;
                    x = mx;
                    break;
                }
            }
            while (last - x > ans) {
                long long mn, mx;
                MinMax(x + 1, x + ans, &mn, &mx);
                if (mn == -1) {
                    break;
                } else {
                    x = mx;
                }
            }
        }

        return ans;
    }
}

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