이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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]);
}
}
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