Submission #1033252

#TimeUsernameProblemLanguageResultExecution timeMemory
1033252GrindMachineChoreography (IOI23_choreography)C++17
100 / 100
63 ms9404 KiB
#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> p,q; int ip[2][2], fp[2][2]; void init(int n_, std::vector<int> p_) { n = n_; p = p_; q = vector<int>(n); rep(i,n) q[p[i]] = i; rep(i,n) a.pb({i,p[i]}); j = 0; } void move_right(int k) { rep(x,2){ ip[j][x] += k; ip[j][x] %= n; } int ofp[2][2]; rep(x,2) rep(y,2) ofp[x][y] = fp[x][y]; rep(x,2){ int prev_parity = (x-k); prev_parity &= 1; fp[j][x] = ofp[j][prev_parity]-k; fp[j][x] = (fp[j][x]%n+n)%n; } } void move_left(int k) { move_right(n-k); } void swap_places() { rep(x,2){ if((x+ip[j][x])&1) ip[j][x]--; else ip[j][x]++; ip[j][x] = (ip[j][x]%n+n)%n; } int ofp[2][2]; rep(x,2) rep(y,2) ofp[x][y] = fp[x][y]; rep(x,2){ int add = 1; if(x) add = -1; fp[j][x] = ofp[j][x^1]+add; fp[j][x] = (fp[j][x]%n+n)%n; } } void move_around() { j ^= 1; } int get_position(int d){ int change = fp[j^1][d&1]; d += change; d %= n; if(j^1){ d = q[d]; } int i = a[d][j]; i += ip[j][i&1]; i %= n; return i; }
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