B. Quick Sort

By xiaruize

思路

按照 1→n1 \rightarrow n 的顺序遍历每个数

因为每次操作是将 kk 个数放在数组末尾， 所以在 xx 上一旦使用这个操作，那么 x−nx-n 中的所有数都需要经过一次操作

所以要最大化 xx ，即求形如 1,2,3⋯1,2,3\cdots 的子序列的最大长度

Code

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/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e5 + 10;

// bool st;
int t;
int n, k;
int a[N];
int pos[N];
// bool en;

void solve()
{
    cin >> n >> k;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        pos[a[i]] = i;
    }
    int x = 1;
    int res = 0;
    for (int i = 1; i < n; i++)
    {
        if (pos[i + 1] > pos[i])//不需要操作
            continue;
        //因为下一位在当前位前面，所以不得不进行操作
        cout << (n - i + k - 1) / k << endl;
        return;
    }
    //递增
    cout << "0" << endl;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> t;
    while (t--)
        solve();
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}