Submission #473280

#	Time	Username	Problem	Language	Result	Execution time	Memory
473280		_L__	Doktor (COCI17_doktor)	C++17	60 / 100	1094 ms	20956 KiB

doktor

// This code is written by _L__
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define F_word ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
typedef long long ll;
typedef long double ld;
const int mod = 1e9+7, N = 2e5+13, inf = 1e9+1;
const ld E = 1e-6;
#define ff first
#define ss second
 
int main(void){
    F_word;
    int n; cin >> n;
    int a[n+2] = {}, doc[n+4] = {}; for(int i = 1; i <= n; ++i){cin >> a[i]; doc[a[i]] = a[i]-i;}a[0] = 1;
    if(n <= 500){
        int sum = 0, ans1 = 1, ans2 = 1;
        for(int i = 1; i <= n; ++i){
            for(int j = i; j <= n; ++j){
                reverse(a+i, a+j+1);
                ll x = 0; for(int k = 1; k <= n; ++k) x += (a[k] == k);
                if(x >= sum){
                    ans1 = i, ans2 = j, sum = x;
                }
                reverse(a+i, a+j+1);
            }
        }
        cout << a[ans1] << ' ' << a[ans2] << endl;
        return 0;
    }
    map<int, int> freq_odd, freq_even;
    for(int i = 1; i <= n; ++i)
        if(!(doc[a[i]]%2))
            freq_odd[(i+a[i])/2]++;
    for(int i = 1; i <= n; ++i)
        if(doc[a[i]]%2)
            freq_even[(i+a[i])/2]++;
    ll ans_odd = 0, tar = 0, ans_even = 0;
    for(auto u: freq_odd)
        if(u.ss >= tar)
            {tar = u.ss, ans_odd = u.ff;}
    tar = 0;
    for(auto u: freq_even)
        if(u.ss >= tar)
            {tar = u.ss, ans_even = u.ff;}
    ll a_1 = ans_odd, a_2 = ans_odd, sum = 0, i = ans_odd, j = ans_odd;
    while(i > 0 && j <= n){
        reverse(a+i, a+j+1); ll x = 0;
        for(int k = 1; k <= n; ++k) x += (a[k] == k);
        if(x >= sum){
            a_1 = i, a_2 = j; sum = x;
        }
        reverse(a+i, a+j+1); ++j, --i;
    }
    ll a_3 = ans_even, a_4 = ans_even+1, sum1 = 0; if(a_3 == 0 || a_4 == 0){a_3 = 0, a_4 = 0;} i = a_3, j = a_4;
    while(i > 0 && j <= n){
        reverse(a+i, a+j+1); ll x = 0;
        for(int k = 1; k <= n; ++k) x += (a[k] == k);
        if(x >= sum1){
            a_3 = i, a_4 = j; sum1 = x;
        }
        reverse(a+i, a+j+1); ++j, --i;
    }
    if(sum >= sum1) cout << a[a_1] << ' ' << a[a_2] << endl;
    else cout << a[a_3] << ' ' << a[a_4] << endl;
} 

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10