Submission #1309108

#	Time	Username	Problem	Language	Result	Execution time	Memory
1309108		minh30082008	Ancient Machine (JOI21_ancient_machine)	C++20	0 / 100	34 ms	6444 KiB

Anna

#include "Anna.h"
//#include "Bruno.h"
#include<bits/stdc++.h>
#define fi first
#define se second
#define FOR(i, k, n) for(int i = k; i <= n; i++)
#define FOR1(i, k, n) for(int i = k; i >= n; i--)
#define pb push_back
#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define vi vector<int>
#define pii pair<int, int>
#define vii vector<pii>
#define ll long long
#define vll vector<ll>
#define pll pair<ll, ll>
#define re return 0
#define mii map<int, int>
#define input "ALONE.inp"
#define output "ALONE.out"
#define rf 	freopen(input, "r", stdin); freopen(output, "w", stdout)
using namespace std;
ll f[100];
void Anna(int n, vector<char> s) {
	int vt = -1;
	FOR(i, 0, n - 1)
		if(s[i] == 'X')
		{
			vt = i;
			break;
		}
	f[0] = 1;
	f[1] = 2;
	FOR(i, 2, 70)
		f[i] = f[i - 1] + f[i - 2];
	vi vv;
	if(vt == -1)
	{
		FOR(i, 0, n - 1)
			vv.pb(0);
	}
	else
	{
		FOR(i, 0, vt - 1)
			vv.pb(0);
		vv.pb(1);
		if(vt + 1 < n)
			vv.pb(0);
		FOR(i, vt + 2, n - 1)
			if(s[i] == 'Z' && s[i - 1] != 'Z')
				vv.pb(1);
			else
				vv.pb(0);
	}
	int l = 0, r = 63;
	while(l < n)
	{
		r = min(r, n - 1);
		ll res = 0;
		FOR(i, l, r)
			if(vv[i])
				res += f[i - l];
		FOR(i, 0, 46)
			Send((res >> i) & 1);
		l = r + 1;
		r += 64;
	}
}

Bruno

//#include "Anna.h"
#include "Bruno.h"
#include<bits/stdc++.h>
#define fi first
#define se second
#define FOR(i, k, n) for(int i = k; i <= n; i++)
#define FOR1(i, k, n) for(int i = k; i >= n; i--)
#define pb push_back
#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define vi vector<int>
#define pii pair<int, int>
#define vii vector<pii>
#define ll long long
#define vll vector<ll>
#define pll pair<ll, ll>
#define re return 0
#define mii map<int, int>
#define input "ALONE.inp"
#define output "ALONE.out"
#define rf 	freopen(input, "r", stdin); freopen(output, "w", stdout)
using namespace std;
ll f[100];
void Bruno(int n, int L, vi s) {
	f[0] = 1;
	f[1] = 2;
	FOR(i, 2, 70)
		f[i] = f[i - 1] + f[i - 2];
	int l = 0, r = 63, d = 0;
	vi s1, tmp;
	while(l < n)
	{
		r = min(r, n - 1);
		ll res = 0;
		FOR(i, 0, 46)
		{
			if(d + i < L && s[d + i])
				res |= (1ll << i);
		}
		tmp.clear();
		FOR1(i, r - l, 0)
			if(res >= f[i])
			{
				res -= f[i];
				tmp.pb(1);
			}
			else
				tmp.pb(0);
		reverse(tmp.begin(), tmp.end());
		for(auto x : tmp)
			s1.pb(x);
		l = r + 1;
		r += 64;
		d += 47;
	}
	int vt = n;
    FOR(i, 0, n - 1)
    {
        if(s1[i])
        {
            vt = i;
            break;
        }
        Remove(i); // Xóa tất cả những gì trước X
    }

    if(vt == n) return; // Không tìm thấy X

    vector<int> vv;
    vv.pb(vt); // vv[0] là X

    FOR(i, vt + 1, n - 1)
    {
        if(s1[i] == 1) // Gặp tín hiệu xóa (Z)
        {
            while(vv.size() > 1) // Xóa các phần tử Y tích lũy, giữ lại X
            {
                Remove(vv.back());
                vv.pop_back();
            }
            Remove(i); // Xóa chính ký tự Z này
        }
        else
        {
            vv.pb(i); // Tích lũy vào stack
        }
    }

    // --- ĐOẠN SỬA ---
    // Xóa các phần tử còn dư trong stack, TRỪ phần tử đầu tiên (X)
    for(size_t i = 0; i < vv.size(); i++)
        Remove(vv[i]);
    // ----------------
    
    return;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 95