제출 #728341

#제출 시각아이디문제언어결과실행 시간메모리
728341sunwukong123Super Dango Maker (JOI22_dango3)C++17
100 / 100
4086 ms1052 KiB
#include "dango3.h" #include <bits/stdc++.h> using namespace std; #define FOR(i,s,e) for(int i = s; i <= (int)e; ++i) #define DEC(i,s,e) for(int i = s; i >= (int)e; --i) #define IAMSPEED ios_base::sync_with_stdio(false); cin.tie(0); #ifdef LOCAL #define db(x) cerr << #x << "=" << x << "\n" #define db2(x, y) cerr << #x << "=" << x << " , " << #y << "=" << y << "\n" #define db3(a,b,c) cerr<<#a<<"="<<a<<","<<#b<<"="<<b<<","<<#c<<"="<<c<<"\n" #define dbv(v) cerr << #v << ":"; for (auto ite : v) cerr << ite << ' '; cerr <<"\n" #define dbvp(v) cerr << #v << ":"; for (auto ite : v) cerr << "{" << ite.f << ',' << ite.s << "} "; cerr << "\n" #define dba(a,ss,ee) cerr << #a << ":"; FOR(ite,ss,ee) cerr << a[ite] << ' '; cerr << "\n" #define reach cerr << "LINE: " << __LINE__ << "\n"; #else #define reach #define db(x) #define db2(x,y) #define db3(a,b,c) #define dbv(v) #define dbvp(v) #define dba(a,ss,ee) #endif mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); #define pb push_back #define eb emplace_back #define all(x) (x).begin(), (x).end() #define f first #define s second #define g0(x) get<0>(x) #define g1(x) get<1>(x) #define g2(x) get<2>(x) #define g3(x) get<3>(x) typedef pair <int, int> pi; typedef tuple<int,int,int> ti3; typedef tuple<int,int,int,int> ti4; int rand(int a, int b) { return a + rng() % (b-a+1); } const int MOD = 1e9 + 7; const int inf = (int)1e9 + 500; const long long oo = (long long)1e18 + 500; template <typename T> void chmax(T& a, const T b) { a=max(a,b); } template <typename T> void chmin(T& a, const T b) { a=min(a,b); } const int MAXN = -1; vector<int> lft; int n,m; int Ans[405]; vector<int> V[405]; vector<int> out; int qqc; int Query_max(vector<int> vec) { int have[n*m+5]; memset(have,0,sizeof have); for(auto x:vec)have[x]=1; vector<int> qq; FOR(i,1,n*m)if(have[i]==0)qq.pb(i); ++qqc; int res=Query(qq); return m-res; } void Solve(int n, int m) { ::n=n; ::m=m; vector<int> vec; FOR(i,1,n*m) { // N*M=1000 int lo=0,hi=m+1; while(lo<hi-1) { // log2(M)= int mid=(lo+hi)/2; vector<int> Q=V[mid]; Q.pb(i); if(Query_max(Q) > 1) lo=mid; else hi=mid; } V[hi].pb(i); } FOR(i,1,m){ Answer(V[i]); } }
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...