[GYM] 2024 Argentinian Programming Tournament (TAP)

3 недели назад, # ^ |

For J

Hint 1

Hint 2

There is more to it than these hints, but IMO the problem is mostly annoying to implement rather than difficult to get to the answer.

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_Shero

2 недели назад, # ^ |

I did this but it still gives me wrong on test 7?!

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2 недели назад, # ^ |

What about numbers that are contiguous and add up to X, just think about all the edge cases.

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MonkeWFS

2 недели назад, # ^ |

Isn't it simply a question of calculating x combinations and seeing if there are enough elements (x-1) to intercede between them?

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13 дней назад, # ^ |

I didn't get what you meant.

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MonkeWFS

10 дней назад, # ^ |

I don't think I ended up solving that one so I don't get what I mean either haha All good!

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benslamaoussama1919

3 недели назад, # |

← Rev. 4 →

could someone please give me a hint for L

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Dani.30

3 недели назад, # ^ |

Try and think if agustin and juan share any numbers

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benslamaoussama1919

3 недели назад, # ^ |

yeah i think about that and augustin will take such a number because he is playing first but still have a wrong answer in test case 8

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6 дней назад, # ^ |

Are you sure the first player will always pick that number? I'd suggest you handle that case separately. Also, keep in mind that, if the amount of times said number appears on the range is odd, one of them will pick it more often.

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ignacioDias

3 недели назад, # |

hint B?

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NanoMonardes

2 недели назад, # ^ |

First of all, it is clear that the usefull segments can be like $$$(1, d)$$$ where $$$d$$$ is a divisor of the number $$$K$$$. Think in cases with a lot of divisors like $$$K = 12$$$, can $$$(1, 2)$$$ or $$$(1, 3)$$$ be minimal or there is another segment that makes them unncesary?

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infinitix

3 недели назад, # |

where can I find the editorial?

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Crisjvm2002

2 недели назад, # |

← Rev. 2 →

Hit C? Plaese

Spoiler

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13 дней назад, # ^ |

That's not enough

Hint

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6 дней назад, # ^ |

Another hint?

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6 дней назад, # ^ |

Sure

Hint 2

Hint 3

Hint 4

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Lacertae

9 дней назад, # |

Can someone give me a hint for E please?

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6 дней назад, # ^ |

← Rev. 2 →

It's a classical DP problem.

Hint 1

Hint 2

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8 дней назад, # |

Hint For Problem L? Please.

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6 дней назад, # ^ |

← Rev. 2 →

Hint 1

Hint 2

Hint 3

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6 дней назад, # ^ |

Thanks

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6 дней назад, # ^ |

I thinks we can we store the count of powers of 2 and count of odd numbers and then ee just compare the sum.

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6 дней назад, # ^ |

You're close.

Hint 4

Hint 5

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6 дней назад, # ^ |

-8

Yah I have used but It gives me wrong answer on test case 4 and this is the code i have used


#include <bits/stdc++.h>
using namespace std;

int main(){
 int n, q;
 cin >> n >> q;
 vector<int> a(n);
 for(int i = 0; i < n; ++i){
  cin >> a[i];
 }
 vector<long long> x(n, 0), y(n, 0);
 if((a[0] & (a[0] - 1)) == 0) x[0] = a[0];
 for(int i = 1; i < n; ++i){
  if((a[i] & (a[i] - 1)) == 0) x[i] += x[i - 1] + a[i];
  else x[i] = x[i-1];
 }
 if(a[0] & 1) y[0] = a[0];
 for(int i = 1; i < n; ++i){
  if(a[i] & 1) y[i] += y[i - 1] + a[i];
  else y[i] = y[i-1];
 }
 while(q--){
  int lo, hi;
  cin >> lo >> hi;
  lo--, hi--;
  if(lo == 0){
   long long f = x[hi];
   long long s = y[hi];
   cout << (f > s ? "A" : f < s ? "B"  : "E") << '\n';
   continue;
  }
  long long f = x[hi] - x[lo - 1];
  long long s = y[hi] - y[lo - 1];
  cout << (f > s ? "A" : f < s ? "B"  : "E") << '\n';
 }
}

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