思路：
自我感觉这题比B简单，它的目的是让连续不同的最小值最大，呢么我们就利用出现最多的当作隔板，看能把这个序列最大分成几段，显然这个中间段就算最大的，不然挤在一起，中间段肯定存在比当前小的。
参考代码：
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
using namespace std;
typedef long long ll;
//#define ll long long
const ll N = 1e3 + 5;
const ll maxn = 1e5 + 20;
const ll mod = 1000000007;
int inv[maxn], vis[maxn], dis[maxn];
int fac[maxn], a[maxn], q[maxn], b[N], c[N];
vector<ll> vec;
//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p> > m;
// ll sum[maxn], a[maxn];
ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
map<ll, ll> mp;
ll ksm(ll a, ll b)
{a %= mod;ll ans = 1ll;while (b){if (b & 1)ans = (ans * a) % mod;a = (a * a) % mod;b >>= 1ll;}return ans;
}
// ll dp[105][16005];
// string p = "abacaba";
// queue<ll> qk, q;
//vector<ll> vec;
// ll sumx[maxn], sumy[maxn], sumk[maxn];
int cnt;
map<int, int> p;
struct node
{int ls, rs, sum;
} tr[maxn * 32];
void inser(int &k, int L, int R, int pos, int w)
{if (!k)k = ++cnt;tr[k].sum += w;if (L == R)return;int mid = L + R >> 1;if (mid >= pos)inser(tr[k].ls, L, mid, pos, w);elseinser(tr[k].rs, mid + 1, R, pos, w);
}
int query(int k, int L, int R, int ik)
{if (L == R)return L;int sk = tr[tr[k].ls].sum;int mid = L + R >> 1;if (sk >= ik)return query(tr[k].ls, L, mid, ik);elsereturn query(tr[k].rs, mid + 1, R, ik - sk);
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);ll t, n, ans;// scanf("%lld", &t);cin >> t;while (t--){p.clear();int n, ans = 0;cin >> n;for (int i = 0; i < n; i++)cin >> a[i], ans = max(++p[a[i]], ans);int k = 0;for (int i = 0; i < n; i++)if (p[a[i]] == ans)k++, p[a[i]] = 0;//cout << ans << ' ' << k << endl;int res = (n - k * ans) / (ans - 1) + k - 1;cout << res << endl;}
}