제출 #728118

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
728118		sunwukong123	Super Dango Maker (JOI22_dango3)	C++17	15 / 100	632 ms	604 KiB

dango3

#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;

vector<int> V[30];
vector<int> out;
bool check(int mid, vector<int> &vec) {
    vector<int> qq;
    for(auto i:out)qq.pb(i);
    FOR(i,0,mid-1)qq.pb(vec[i]);
    int res=Query(qq);
    return !!res;
}
void Solve(int n, int m) {
    ::n=n;
    ::m=m;
    vector<int> vec;
    FOR(i,1,n*m)vec.pb(i);
    FOR(i,1,m) {
        vector<int>nxt;
       
        out.pb(vec.back());
        vec.pop_back();
        // there needs to be n-1 things here.
        FOR(ti,1,n-1) {
            int lo=0,hi=vec.size();
            while(lo<hi-1) {
                int mid=(lo+hi)/2;
                if(check(mid,vec)){ // if i exclude everything from the back until mid.
                    hi=mid;
                } else {
                    lo=mid;
                }
            }
            // my element is vec[hi-1];
            while((int)vec.size()>lo+1){
                nxt.pb(vec.back());
                vec.pop_back();
            }
            int x=vec.back();
            out.pb(x);
            vec.pop_back();
        }
        Answer(out);
        out.clear();
        vec=nxt;
    }
}

• Subtask 1: 0 / 2
• Subtask 2: 0 / 5
• Subtask 3: 15 / 15
• Subtask 4: 0 / 78