제출 #564253

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
564253		Ooops_sorry	Gap (APIO16_gap)	C++14	0 / 100	72 ms	1144 KiB

gap

#include<bits/stdc++.h>
#ifndef LOCAL
    #include "gap.h"
#endif

#define ll long long

using namespace std;

#ifdef LOCAL
    static void my_assert(int k){ if (!k) exit(1); }

    static int subtask_num, N;
    static long long A[100001];
    static long long call_count;
    void MinMax(long long s, long long t, long long *mn, long long *mx)
    {
        int lo = 1, hi = N, left = N+1, right = 0;
        my_assert(s <= t && mn != NULL && mx != NULL);
        while (lo <= hi){
            int mid = (lo+hi)>>1;
            if (A[mid] >= s) hi = mid - 1, left = mid;
            else lo = mid + 1;
        }
        lo = 1, hi = N;
        while (lo <= hi){
            int mid = (lo+hi)>>1;
            if (A[mid] <= t) lo = mid + 1, right = mid;
            else hi = mid - 1;
        }
        if (left > right) *mn = *mx = -1;
        else{
            *mn = A[left];
            *mx = A[right];
        }
        if (subtask_num == 1) call_count++;
        else if (subtask_num == 2) call_count += right-left+2;
    }
#endif // LOCAL

set<ll> st;
const ll INF = 1e18;

mt19937_64 rnd(51);

long long findGap(int t, int n) {
    ll a, b;
    MinMax(0, INF, &a, &b);
    ll len = (b - a - 1) / (n - 1), now = a + 1, last = a, ans = 0;
    for (int i = 0; i < n; i++) {
        ll mn, mx;
        MinMax(now, now + len - 1, &mn, &mx);
        if (mn != -1) {
            if (last != -1) {
                ans = max(ans, mn - last);
            }
            last = mx;
        }
        now += len;
    }
    return ans;
}

#ifdef LOCAL
    int main()
    {
        FILE *in = stdin, *out = stdout;
        my_assert(2 == fscanf(in, "%d%d", &subtask_num, &N));
        my_assert(1 <= subtask_num && subtask_num <= 2);
        my_assert(2 <= N && N <= 100000);
        for (int i=1;i<=N;i++) my_assert(1 == fscanf(in, "%lld", A+i));
        for (int i=1;i<N;i++) my_assert(A[i] < A[i+1]);
        fprintf(out, "%lld\n", findGap(subtask_num, N));
        fprintf(out, "%lld\n", call_count);
    }

#endif

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