2
2
1 10
3
2 5 4

Case #1:
0
Case #2:
1

二分答案，每一个数贪最小

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<queue>
#include<vector>
#include<deque>
#include<complex>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int a[MAXN],n;
bool is_ok(int c)
{
	int p=a[1]-c;
	Fork(i,2,n)
	{
		if (a[i]>p)
		{
			p=max(a[i]-c,p+1);
		}
		else
		{
			if (a[i]+c<=p) return 0;
			else p=min(a[i]+c,p+1);
		}
	}	
	
	
	
	return 1;
}
int main()
{
//	freopen("seq.in","r",stdin);
//	freopen(".out","w",stdout);
	
	int T;
	cin>>T;
	For(kcase,T)
	{
		cin>>n;
		For(i,n) scanf("%d",&a[i]); 
		
		ll l=0,r=F,ans=r,m=-1;
		while (l<=r)
		{
			m=(l+r)/2;
			if (!is_ok(m)) l=m+1;
			else r=m-1,ans=m; 
		}
		
		
		
		printf("Case #%d:\n",kcase);
		cout<<ans<<endl;
	}
	
	
	
	return 0;
}