Submission #1033237

#	Time	Username	Problem	Language	Result	Execution time	Memory
1033237		GrindMachine	Choreography (IOI23_choreography)	C++17	0 / 100	1058 ms	5068 KiB

choreography

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*

already knew the key solution idea

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "choreography.h"

vector<array<int,2>> a;
int j;
int n;
vector<int> delta(2);
vector<int> p,q;

void init(int n_, std::vector<int> p_) {
    p = p_;
    q = vector<int>(n);
    rep(i,n) q[p[i]] = i;
    n = n_;
    rep(i,n) a.pb({i,p[i]});
    j = 0;
} 

void move_right(int k) {
    delta[j] += k;
    delta[j] %= n;
}

void move_left(int k) {
    move_right(n-k);
}

void swap_places() {
    rep(i,n){
        a[i][j] ^= 1;
    }
}

void move_around() {
    j ^= 1;
}

int get_position(int d){
    d -= delta[j^1];
    d = (d%n+n)%n;
    
    rep(i,n){
        if(a[i][j^1] == d){
            int ind = a[i][j];
            return (ind+delta[j])%n;
        }
    }

    assert(0);

    // int i = a[d][j^1];
    // i += delta[j];
    // i %= n;

    // return i;
}

• Subtask 1: 0 / 7
• Subtask 2: 0 / 12
• Subtask 3: 0 / 10
• Subtask 4: 0 / 14
• Subtask 5: 0 / 28
• Subtask 6: 0 / 29