B. Arrays Sum

// Problem: B. Arrays Sum
// Contest: Codeforces - Grakn Forces 2020
// URL: https://codeforces.com/problemset/problem/1408/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-02-25 14:03:18
// 
// Powered by CP Editor (https://cpeditor.org)

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

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
void solve() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n + 1);
    
    for (int i = 1;  i<= n ;i ++)
    {
        cin>> a[i];           
    }
    int cnt = 1;
    for (int i = 2; i <= n ;i ++)
        if(a[i] != a[i - 1]) cnt ++;
    if (k == 1) {
        if (cnt == 1) cout <<"1" << endl;
        else cout << "-1" << endl;
        return;
    }
    cout << max(1, ((cnt - 1 + k - 2) / (k - 1))) << endl;
    //上取整转换
}
 
int main () {
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

ceil(x / c) = = (x + c - 1) / c; 这种题需要知道对于1的，才会出现无解。否则一直有解 这样想。 对于我们预处理一个数组na = {a[2] - a[1], a[3] - a[2], ....} 我们对于一个b[i] = {b[i][2] - b[i][1], ....} 至多只有k - 1个非0数，如果na有c个非0，那么我们就至少需要ceil(c / (k - 1)) 我一直想通过递推找到一个具体的解。但是陷入了循环。不够就要加不够就要加。这样就混了