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 <cmath>
#include <cstdio>
#include <algorithm>
#include <random>
#include <string>
#include <bitset>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <list>
#include <set>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <typeinfo>

#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define FORR(i, m, n) for(int i = m; i >= n; i--)
#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define llong long long
#define pb(a) push_back(a)
#define INF 999999999

using namespace std;

typedef pair<int, int> P;
typedef pair<llong, llong> LP;
typedef pair<int, P> PP;
typedef pair<llong, LP> LPP;

int dy[] = {0, 0, 1, -1, 0};
int dx[] = {1, -1, 0, 0, 0};

llong pow(int x, int n){
    llong ans = x;
    if (n == 0) return 1;

    for(int i = 0; i < n - 1; i++) {
        ans *= x;
    }

    return ans;
}

template <typename T> struct BIT { // 1-indexed
private:
    vector<T> array;
    const int n;

public:
    // 初期化
    BIT(int _n) : array(_n + 1, 0), n(_n) {}

    // 1番目からi番目までの累積和を求める
    T sum(int i) {
        T s = 0;
        while(i > 0) {
            s += array[i];
            i -= i & -i; // LSB減算
        }

        return s;
    }

    // [i, j]の要素の総和
    T sum(int i, int j) {
        T ret_i = sum(i - 1);
        T ret_j = sum(j);

        return ret_j - ret_i;
    }

    // i番目に要素xを追加
    void add(int i, T x) {
        while (i <= n) {
            array[i] += x;
            i += i & -i; // LSB加算
        }
    }
};

template <typename T> struct BIT2 { // 1-indexed
private:
    vector<vector<T> > array;
    const int n;
    const int m;

public:
    // 初期化
    BIT2(int _n, int _m) : array(_n + 1, vector<T>(_m + 1, 0)), n(_n), m(_m) {}

    // (1, 1)から(x, y)までの累積和を求める
    T sum(int x, int y) {
        T s = 0;
        for (int i = x; i > 0; i -= i & (-i)) {
            for (int j = y; j > 0; j -= j & (-j)) {
                s += array[i][j];
            }
        }

        return s;
    }

    // [(x1, y1), (x2, y2)]の要素の総和
    T sum(int x1, int y1, int x2, int y2) {
        return sum(x2, y2) - sum(x1 - 1, y2) - sum(x2, y1 - 1) + sum(x1 - 1, y1 - 1);
    }

    // (x, y)に要素kを追加
    void add(int x, int y, T k) {
        for (int i = x; i <= n; i += i & (-i)) {
            for (int j = y; j <= m; j += j & (-j)) {
                array[i][j] += k;
            }
        }
    }
};

void swap(int *X, int *Y){
    int t = *X; *X = *Y; *Y = t;
}

string toUpper(string s) {
    string t = s;
    REP(i, s.size()) {
        t[i] = toupper(s[i]);
    }
    return t;
}

string toLower(string s) {
    string t = s;
    REP(i, s.size()) {
        t[i] = tolower(s[i]);
    }
    return t;
}

bool isOdd(llong i) {
    if(i % 2) {
        return true;
    } else {
        return false;
    }
}

int vector_finder(std::vector<string > vec, string number) {
    auto itr = std::find(vec.begin(), vec.end(), number);
    size_t index = std::distance( vec.begin(), itr );
    if (index != vec.size()) { // 発見できたとき
        return 1;
    }
    else { // 発見できなかったとき
        return 0;
    }
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, k;
    cin >> n >> k;
    if (n == k) {
        cout << 1 << endl;
        return 0;
    }
    int tmp = 0;
    for (int i = 0; i < n; ++i) {
        cin >> tmp;
        if (tmp == 1) {
            // cout << "i :";
            // cout << i << endl;
            // cout << "left :";
            // cout << "num :";
            // cout << i << endl;
            // cout << i / (k - 1) << endl;
            // cout << ceil(i / (k - 1)) << endl;
            // cout << "right :";
            // cout << "num :";
            // cout << n - i - 1 << endl;
            // cout << (n - i - 1) / (k - 1) << endl;
            // cout << ceil((n - i - 1) / (k - 1)) << endl;
            cout << ceil((double)(i /(k - 1))) + ceil((double)((n - i - 1) / (k - 1))) << endl;
            break;
        }
    }

    return 0;
}