[Niuke] [noip2015] jumping stone

Preface

$tag:$ Two points Two points answer
The question :

Ideas :
Because what the title asks is Shortest maximum , A very standard two-point inquiry method

So we can consider Two point shortest distance , So how do we check Well

It's simple , Obviously, what needs to be judged now x, If the difference between two stones is less than x When , We must remove it

So we just need to judge the quantity and m The relationship between

Of course, we can move the state through last To transfer
code :

int L,n,m;

int a[N];
int i,j;
int b[N];

bool check(int x){
    
	int cnt = 0;
	int last = 0 ;
	
	Fup(i,1,n+1){
    
	 	if(a[i] - last < x) cnt++;
	 	else last = a[i];
	 	if(cnt > m) return false;
	 	
	}
	return true;
}
void solve(){
    
	// cin>>L>>N>>M;
	cin>>L>>n>>m;
	Fup(i,1,n) cin>>a[i];
	
	a[n+1] = L;
	
	
	int l = 0 , r =  L;
	
	
	
	//cout<<ans<<endl;
	
	
	while(l<=r){
    
		int mid = (l+r)>>1;
		if(check(mid)){
     
			l = mid+1;
		}
		else r = mid - 1;
	}
	
	cout<<((l+r)>>1)<<endl;
	
	
}
signed main(){
    
	//int t;cin>>t;while(t--)
    solve();
    return 0 ;
}