제출 #395798

#제출 시각아이디문제언어결과실행 시간메모리
395798ignaciocantaRabbit Carrot (LMIO19_triusis)C++14
100 / 100
31 ms4500 KiB
#include <bits/stdc++.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+7; const int mod = 998244353; const int MX = 805; 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 main(){ NACHO(); // en vez de buscar minimizar las que cambiamos, // busquemos maximizar las que no cambiamos. // sean a_i1, a_i2, ..., a_ik los palos que no recorto. // claramente, a_ij <= M*ij // y ademas a_ij+1 <= a_ij+M*(ij+1-ij) // de esto sale a_i1 <= M*i1 ---> 0 <= M*i1-ai1 // si sumamos, M*ij, // M*ij+a_ij+1 <= a_ij+M*(ij+1-ij)+M*(ij) // M*ij-aij <= M*ij+1-a_ij+1 // sea bi = M*i-ai // sigue que 0 <= bi <= bi+1 // queremos encontrar entonces la longest non decreasing subsequence. // tadaa int n, M; cin >> n >> M; vi b; F0R(i, n){ int x; cin >> x; if(x <= M*(i+1)){ b.pb(M*(i+1)-x); } } vi dp; F0R(i, sz(b)){ int ind = upper_bound(all(dp), b[i])-dp.begin(); if(ind == sz(dp)) dp.pb(b[i]); else dp[ind] = b[i]; } cout << n-sz(dp) << "\n"; }

컴파일 시 표준 에러 (stderr) 메시지

triusis.cpp: In function 'void NACHO(std::string)':
triusis.cpp:69:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   69 |         freopen((name+".in").c_str(), "r", stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
triusis.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+".out").c_str(), "w", stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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