Submission #220189

#	Time	Username	Problem	Language	Result	Execution time	Memory
220189		rama_pang	JOIRIS (JOI16_joiris)	C++14	15 / 100	6 ms	384 KiB

joiris

#include <iostream>
#include <vector>
using namespace std;

int main() {
  int N, K;
  cin >> N >> K;

  vector<int> A(N);

  for (int i = 0; i < N; i++) {
    cin >> A[i];
  } 

  { // Check existence of solution
    vector<int> B(K);
    for (int i = 0; i < N; i++) {
      B[i % K] += A[i];
    }
    for (int i = 0; i < N % K; i++) {
      if (B[i] != B[0]) {
        cout << -1 << "\n";
        return 0;
      }
    }
    for (int i = N % K; i < K; i++) {
      if (B[i] != B[N % K]) {
        cout << -1 << "\n";
        return 0;
      }
    }
  }

  vector<pair<int, int>> operations; // (type (0 vertical, 1 horizontal), leftmost column)

  { // Step 1: make A[0] <= A[1] <= ... <= A[N - 1] holds
    for (int i = 1; i < N; i++) {
      while (A[i - 1] > A[i]) {
        A[i] += K;
        operations.emplace_back(0, i);
      }
    }

    for (int i = N - 1; i >= 0; i--) {
      A[i] -= A[0]; // delete rows
    }
  }

  { // Step 2: Fill columns K - 1, K, ..., N - 1 with horizontal pieces
    for (int i = 1; i <= A[N - 1]; i++) {
      for (int j = N - 1; j >= 0; j--) {
        if (j + K - 1 < N && A[j + K - 1] < i) {
          operations.emplace_back(1, j);
          for (int k = j; k < j + K; k++) {
            A[k]++;
          }
        }
      }
    }
  }

  { // Step 3: Clear columns K - 1, K, ..., N - 1 by filling 0...K-2 with vertical pieces
    for (int i = 0; i < K - 1; i++) {
      while (A[i] < A[N - 1]) {
        operations.emplace_back(0, i);
        A[i] += K;
      }
    }
    for (int i = 0; i < N; i++) {
      A[i] -= A[N - 1]; // delete rows
    }
  }

  { // Step 4: Fill vertical pieces for A[0], ..., A[N % K - 1] so they are all the same.
    int mxheight = 0;
    for (int i = 0; i < N % K; i++) {
      mxheight = max(mxheight, A[i]);
    }
    for (int i = 0; i < N % K; i++) {
      while (A[i] < mxheight) {
        operations.emplace_back(0, i);
        A[i] += K;
      }
    }
  }

  { // Step 5: Fill vertical pieces for A[N % K], ..., A[N - 1] so they are all the same
    int mxheight = 0;
    for (int i = N % K; i < N; i++) {
      mxheight = max(mxheight, A[i]);
    }
    for (int i = N % K; i < N; i++) {
      while (A[i] < mxheight) {
        operations.emplace_back(0, i);
        A[i] += K;
      }
    }
  }

  { // Step 6: Make 0...N-1 the same height by adding horizontal positions
    int key = 0;
    for (int i = 0; i < N % K; i++) {
      while (A[i] < A[N % K]) {
        operations.emplace_back(0, i);
        A[i] += K;
      }
      key = A[i];
    }
    for (int i = N % K; i < N; i += K) {
      while (A[i] < key) {
        operations.emplace_back(1, i);
        for (int j = i; j < i + K; j++) {
          A[j]++;
        }
      }
    }
  }

  cout << operations.size() << "\n";
  for (auto &i : operations) {
    cout << i.first + 1 << " " << i.second + 1 << "\n";
  }
  return 0;
}

• Subtask 1: 0 / 15
• Subtask 2: 15 / 15
• Subtask 3: 0 / 15
• Subtask 4: 0 / 55