题意： 有一组数据，你可以任意顺序拿去，求相邻两个相同数字的最小距离的max。
思路：
- 要想最小距离最大，那么意味着需要让相同的数字在整个数组中等距放置。
- 统计得到最大出现次数maxx，再统计出现maxx次的元素个数cnt。详细步骤见代码！
代码实现：
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
int t, n, maxx, cnt, a[100010];
map<int, int> mp;
signed main()
{
IOS;
cin>>t;
while(t --){
cin >> n;
mp.clear();
maxx = 0, cnt = 0;
for(int i = 1; i <= n; i ++){
cin >> a[i], mp[a[i]] ++;
maxx = max(maxx,mp[a[i]]);
}
for(auto it : mp) if(it.second == maxx) cnt ++;
if(maxx == n) cout << 0 << endl;
else if(maxx == 2) cout << n - cnt - 1 << endl;
else{
int tmp = n/maxx;
int m = n%maxx + n/maxx;
m -= cnt;
cout << tmp - 1 + (m / (maxx-1)) << endl;
}
}
return 0;
}