CF-1016-E. Min Max MEX · znzryb

## 思路讲解

非常明显的二分答案，**最小的最大**

## AC代码

[https://codeforces.com/contest/2093/submission/314861882](https://codeforces.com/contest/2093/submission/314861882)



::: 源代码

```cpp
// Problem: E. Min Max MEX
// Contest: Codeforces - Codeforces Round 1016 (Div. 3)
// URL: https://codeforces.com/contest/2093/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(5e18)+3;
constexpr ll range=static_cast<ll>(2e5)+10;

ll N,K,T,A[MAXN];

inline bool check(ll x){
	// 判断mex==x是否可能
	ll cnt=0;
	set<ll> mex;
	FOR(i,1,N){
		if(A[i]<x) mex.insert(A[i]);
		if(mex.size()>=x){
			mex.clear();
			++cnt;
		}
	}
	if(cnt>=K){
		return true;
	}
	return false;
}

inline ll bsearch(ll l,ll r){
	while(l<r){
		ll mid=l+r+1>>1;
		if(check(mid)){
			l=mid;
		}else{
			r=mid-1;
		}
	}
	return l;
}

inline void solve(){
	cin>>N>>K;
	// 将数列分成k组，使最小的mex（bi）最大
	// 最大值最小，有点像二分答案那
	FOR(i,1,N){
		cin>>A[i];
	}
	cout<<bsearch(0,N/K+5)<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		solve();
	}
	// solve();
	return 0;
}
/*
AC
https://codeforces.com/contest/2093/submission/314861882
*/
```


:::


## 心路历程（WA，TLE，MLE……）