Submission #381790

#TimeUsernameProblemLanguageResultExecution timeMemory
381790ignaciocantaGondola (IOI14_gondola)C++14
60 / 100
98 ms13284 KiB
#include <bits/stdc++.h> #include "gondola.h" using namespace std; using tint = long long; using ld = long double; #define FOR(i,a,b) for (int i = (a); i < (b); ++i) #define F0R(i,a) FOR(i,0,a) #define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i) #define R0F(i,a) ROF(i,0,a) #define trav(a,x) for (auto& a: x) using pi = pair<int,int>; using pl = pair<tint,tint>; using vi = vector<int>; using vpi = vector<pi>; using vpl = vector<pl>; using vvi = vector<vi>; using vl = vector<tint>; using vb = vector<bool>; #define pb push_back #define pf push_front #define rsz resize #define all(x) begin(x), end(x) #define rall(x) x.rbegin(), x.rend() #define sz(x) (int)(x).size() #define ins insert #define f first #define s second #define mp make_pair #define DBG(x) cerr << #x << " = " << x << endl; const int MOD = 1e9+9; //change this const tint mod = 998244353; const int MX = 5005; const tint INF = 1e18; const int inf = 2e9; const ld PI = acos(ld(-1)); const ld eps = 1e-5; const int dx[4] = {1, -1, 0, 0}; const int dy[4] = {0, 0, 1, -1}; template<class T> void remDup(vector<T> &v){ sort(all(v)); v.erase(unique(all(v)),end(v)); } template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; } template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; } bool valid(int x, int y, int n, int m){ return (0<=x && x<n && 0<=y && y<m); } int cdiv(int a, int b) { return a/b+((a^b)>0&&a%b); } //redondea p arriba int fdiv(int a, int b) { return a/b-((a^b)<0&&a%b); } //redondea p abajo void NACHO(string name = ""){ ios_base::sync_with_stdio(0); cin.tie(0); if(sz(name)){ freopen((name+".in").c_str(), "r", stdin); freopen((name+".out").c_str(), "w", stdout); } } int valid(int n, int b[]) { // si algun numero aparece dos veces, no se puede. // si hay dos numeros <= n (no se cambiaron), hay // que chequear que sean validos // si son todos >= n o todos excepto 1, no hay forma de verificar nada. vi id (n); vi a (n); F0R(i, n) a[i] = b[i]; F0R(i, n) id[i] = i+1; vi c (250005, 0); F0R(i, n) ++c[a[i]]; F0R(i, 250005) if(c[i] > 1) return 0; int pos = -1; F0R(i, n) if(a[i] <= n) pos = i; if(pos == -1) return 1; rotate(id.begin(), id.begin()+a[pos]-1, id.end()); rotate(a.begin(), a.begin()+pos, a.end()); F0R(i, n) if(a[i] <= n && a[i] != id[i]) return 0; return 1; } //---------------------- int replacement(int n, int b[], int replacementSeq[]) { bool allG = 1; vi id (n); vi a (n); F0R(i, n) a[i] = b[i]; F0R(i, n) id[i] = i+1; F0R(i, n) if(a[i] <= n) allG = 0; vi c (n); F0R(i, n) c[i] = a[i]; if(allG){ F0R(i, n) a[i] = i+1; } // uso el pivot para saber que gondola cambiada se corresponde // con que gondola original int pos; F0R(i, n) if(a[i] <= n) pos = i; rotate(id.begin(), id.begin()+a[pos]-1, id.end()); rotate(a.begin(), a.begin()+pos, a.end()); rotate(c.begin(), c.begin()+pos, c.end()); vpi g; if(allG) F0R(i, n) a[i] = c[i]; F0R(i, n){ if(a[i] > n) g.pb(mp(a[i], i)); } if(sz(g) == 0) return 0; sort(all(g)); int ind = 0; set<int> av; FOR(i, n+1, 250005) av.ins(i); F0R(i, n) if(a[i] > n) av.erase(a[i]); F0R(i, sz(g)){ int cur = id[g[i].s]; while(cur < g[i].f){ replacementSeq[ind] = cur; cur = *av.begin(); av.erase(*av.begin()); ++ind; } av.ins(cur); } return ind; } //---------------------- int add(int a, int b){ return (a+b) % MOD; } int sub(int a, int b){ return (a-b+MOD) % MOD; } int mul(int a, int b){ return (tint)a * b % MOD; } int bexp(int a, int b){ int ret = 1; a %= MOD; while(b > 0){ if(b & 1) ret = mul(ret, a); a = mul(a, a); b /= 2; } return ret; } int countReplacement(int n, int b[]) { if(!valid(n, b)) return 0; // sea maxi el maximo numero de gondola // claramente, tengo que insertar // todas las gondolas entre [n+1;maxi] para // llegar a maxi. // ahora, cuantas formas tengo de poner la gondola x? // puedo ponerla en la cantidad de lugares cuya gondola final // es mayor a x. // sin embargo, si vas de n+1 a maxi es a lo sumo 1e9, por lo cual esto // no es viable. // lo que hay que notar es lo siguiente. // supongamos que tenemos ... ... ... 12 29 36 // para todos los numeros entre [12;28], la respuesta es 2. // por lo tanto, no hace falta iterar poor todos ellos // ya que su respuesta es la misma (podemos hacer 2^(29-12). bool allG = 1; F0R(i, n) if(b[i] <= n) allG = 0; vi a; F0R(i, n){ if(b[i] > n) a.pb(b[i]); } a.pb(n+1); sort(all(a)); int ret = 1; FOR(i, 1, sz(a)){ ret = mul(ret, bexp(sz(a)-i, a[i]-a[i-1])); } if(allG) ret = mul(ret, n); return ret; }

Compilation message (stderr)

gondola.cpp: In function 'void NACHO(std::string)':
gondola.cpp:70:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   70 |         freopen((name+".in").c_str(), "r", stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
gondola.cpp:71:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   71 |         freopen((name+".out").c_str(), "w", stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
gondola.cpp: In function 'int replacement(int, int*, int*)':
gondola.cpp:116:37: warning: 'pos' may be used uninitialized in this function [-Wmaybe-uninitialized]
  116 |  rotate(id.begin(), id.begin()+a[pos]-1, id.end());
      |                                     ^
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