oj.uz

답안 #778850

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
778850 2023-07-10T20:52:32 Z epicci23 Red-blue table (IZhO19_stones) C++17
0 / 100
 49 ms 2308 KB 
#include "bits/stdc++.h"
#pragma optimize ("Bismillahirrahmanirrahim")
using namespace std;
#define pb push_back
#define ff first
#define ss second
#define endl "\n" 
#define int long long
#define double long double
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define what_is(x) cerr << #x << " is " << x << endl;
//#define m (l+r)/2
constexpr int N=200005;
constexpr int MOD=1000000007;
constexpr int  INF2 = LLONG_MAX;
constexpr int INF=(int)1e18;
constexpr int LOG=30;
typedef pair<int,int> pii;
typedef tuple<int,int,int> tp;
typedef priority_queue<pii,vector<pii>,greater<pii>> min_pq;
typedef priority_queue<pii> max_pq;
typedef long long ll;
//to think//
/*
 * graph approach
 * dp
 * dividing the problem to smaller statements
 * finding the real constraint
 * sqrt decomposition
 * greedy approach
 * pigeonhole principle
 * rewriting the problem/equality 
 * bitwise approaches
 * binary search if monotonic
 * divide and conquer
 * combinatorics
 * inclusion - exclusion
 * think like bfs
*/



inline int in()
{
  int x;cin >> x;
  return x;
}

inline string in2()
{
  string x;cin >> x;
  return x;
}


void solve()
{
  int n=in(),m=in();
  int ans=0,ans2=0;

  char ar[n+1][m+1];
  char ar2[n+1][m+1];

  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
      {ar[i][j]='-';ar2[i][j]='+';}


  int kac=0;
  int cur=m;ans=cur;
  int l=0,xm=0,r=0;
  int uz[3];
  uz[1]=(m&1) ? 1 : 2;
  uz[0]=uz[2]=(m-uz[1])/2;

  bool ok=1;
  

  for(int i=1;i<=n;i++)
  {
    xm++;cur++;
    if(xm==(n+1)/2) cur-=uz[1];
    
    if(ok)
    {
      l++; 
      if(l==(n+1)/2) cur-=uz[0];
    }
    else
    {
      r++;
      if(r==(n+1)/2) cur-=uz[2];
    }
    
    if(cur>ans)
    {
      ans=cur;
      kac=i;
    }

    ok^=1;
  }
 
  ok=1;
  //cout << kac << endl;
  for(int i=1;i<=kac;i++)
  {
    if(ok)
      for(int j=1;j<=m/2+1;j++) ar[i][j]='+';
    else
      for(int j=m;j>=m-(m/2+1)+1;j--) ar[i][j]='+';
    ok^=1;
  }
 

  
  int cur2=n;ans2=cur2;
  l=0,xm=0,r=0,kac=0;
  uz[1]=(n&1) ? 1 : 2;
  uz[0]=uz[2]=(n-uz[1])/2;
  ok=1;
  
  //what_is(cur2);
  for(int i=1;i<=m;i++)
  {
    xm++;cur2++;
    if(xm==(m+1)/2) cur2-=uz[1];
    
    if(ok)
    {
      l++; 
      if(l==(m+1)/2) cur2-=uz[0];
    }
    else
    {
      r++;
      if(r==(m+1)/2) cur2-=uz[2];
    }
    
    //what_is(cur2);

    if(cur2>ans2)
    {
      ans2=cur2;
      kac=i;
    }

    ok^=1;
  }
 
  ok=1;

  for(int i=1;i<=kac;i++)
  {
    if(ok)
      for(int j=1;j<=n/2+1;j++) ar2[j][i]='-';
    else
      for(int j=n;j>=n-(n/2+1)+1;j--) ar2[j][i]='-';
    ok^=1;
  }

  
  cout << max(ans,ans2) << endl;

  if(ans>ans2)
  {
    for(int i=1;i<=n;i++)
      for(int j=1;j<=m;j++)
        cout << ar[i][j] << " \n"[j==m];
  }
  else
  {
    for(int i=1;i<=n;i++)
      for(int j=1;j<=m;j++)
        cout << ar2[i][j] << " \n"[j==m];
  }
}

int32_t main(){
   

     cin.tie(0); ios::sync_with_stdio(0);
     cout << fixed <<  setprecision(15);
   
   int t=1;cin>> t;
 
 for(int i=1;i<=t;i++)
 {
  //  cout << "Case #" << i << ": ";
    solve();
 }
 
 return 0;
}

Compilation message

stones.cpp:2: warning: ignoring '#pragma optimize ' [-Wunknown-pragmas]
    2 | #pragma optimize ("Bismillahirrahmanirrahim")
      |

Subtask #1
0 / 17.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 212 KB Wrong answer
2 Halted 0 ms 0 KB -

Subtask #2
0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 340 KB Wrong answer
2 Halted 0 ms 0 KB -

Subtask #3
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 212 KB Wrong answer
2 Halted 0 ms 0 KB -

Subtask #4
0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 34 ms 2308 KB Wrong answer
2 Halted 0 ms 0 KB -

Subtask #5
0 / 15.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 49 ms 2196 KB Wrong answer
2 Halted 0 ms 0 KB -

Subtask #6
0 / 31.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 212 KB Wrong answer
2 Halted 0 ms 0 KB -