Problem:
Score : 300 points

Problem Statement

There is a sequence of length N: A_1, A_2, ..., A_N. Initially, this sequence is a permutation of 1, 2, ..., N.

On this sequence, Snuke can perform the following operation:

Choose K consecutive elements in the sequence. Then, replace the value of each chosen element with the minimum value among the chosen elements.

Snuke would like to make all the elements in this sequence equal by repeating the operation above some number of times.
Find the minimum number of operations required.
It can be proved that, Under the constraints of this problem, this objective is always achievable.

Constraints

2 \leq K \leq N \leq 100000

A_1, A_2, ..., A_N is a permutation of 1, 2, ..., N.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N

Output

Print the minimum number of operations required.

Sample Input 1

4 3
2 3 1 4

Sample Output 1

2

One optimal strategy is as follows:

In the first operation, choose the first, second and third elements. The sequence A becomes 1, 1, 1, 4.

In the second operation, choose the second, third and fourth elements. The sequence A becomes 1, 1, 1, 1.

Sample Input 2

3 3
1 2 3

Sample Output 2

1

Sample Input 3

8 3
7 3 1 8 4 6 2 5

Sample Output 3

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
#define MAX 100000

void minimization()
{
  int n, k;
  cin >> n >> k;

  cout << ceil((float)(n - 1) / (float)(k - 1)) << "\n";
}

int main()
{

  int N, K;
  scanf("%d %d", &N, &K);

  int A[N], minNum = MAX + 1;
  vector<int> minIndexs;

  // for (int i = 0; i < N; i++)
  // {
  //   scanf("%d", &A[i]);
  // }
  // int g;
  // int a = (N - K) % (K - 1);
  // if (a == 0){
  //   g = (N - K) / (K - 1) + 1;
  // } else {
  //   g = (N - K) / (K - 1) + 2;
  // }

  // cout << g << endl;

  for (int i = 0; i < N; i++)
  {
    scanf("%d", &A[i]);
    if (A[i] < minNum)
    {
      minNum = A[i];
      minIndexs.clear();
    }
    if (A[i] == minNum)
    {
      minIndexs.push_back(i);
    }
  }

  int cnt = 0;
  // headは常に最小値
  int minHead = 0;
  int head = minIndexs[minHead];

  if (head > 0)
  // 左端から埋める
  {
    if ((head + 1) >= K)
    {
      cnt += ceil(head / (K - 1));
      if ((head) % (K - 1) != 0)
      {
        head += (K - 1) - (head) % (K - 1);
      }
    }
    else
    {
      cnt++;
      head = K - 1;
    }
    // 処理済の最小値インデックスを更新
  }
  else
  {
    head += K - 1;
    cnt++;
  }
  minHead++;

  while (head + 1 < N)
  {
    if (minHead + 1 <= minIndexs.size())
    {
      // 次のK個に最小値を含まない場合
      if ((head + K) < minIndexs[minHead])
      {
        // headの最小値を含むK個を置き換えるので、+=K-1
        head += K - 1;
        cnt++;
      }
      else
      {
        // headの最小値を含まないK個を置き換えるので、+=K
        head += K;
        cnt++;
        minHead++;
      }
    }
    else
    {
      // headの最小値を含むK個を置き換えるので、+=K-1
      head += K - 1;
      cnt++;
    }
  }

  cout << cnt;

  return 0;
}