Time Limit: 3000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

LightOJ - 1425


The Monkey and the Oiled Bamboo



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Description




It's time to remember the disastrous moment of the old school math. Yes, the little math problem with the monkey climbing on an oiled bamboo. It goes like:

"A monkey is trying to reach the top of an oiled bamboo. When he climbs up 3 feet, he slips down 2 feet. Climbing up 3 feet takes 3 seconds. Slipping down 2 feet takes 1 second. If the pole is 12 feet tall, how much time does the monkey need to reach the top?"

When I was given the problem, I took it seriously. But after a while I was thinking of killing the monkey instead of doing the horrible math! I had rather different plans (!) for the man who oiled the bamboo.

Now we, the problem-setters, got a similar oiled bamboo. So, we thought we could do better than the traditional monkey. So, I tried first. I jumped and climbed up 3.5 feet (better than the monkey! Huh!) But in the very next second I just slipped and fell off to the ground. I couldn't remember anything after that, when I woke up, I found myself in a bed and the anxious faces of the problem setters around me. So, like old school times, the monkey won with the oiled bamboo.

So, I made another plan (somehow I want to beat the monkey), I took a ladder instead of the bamboo. Initially I am on the ground. In each jump I can jump from the current rung (or the ground) to the next rung only (can't skip rungs). Initially I set my strength factor k. The meaning of k is, in any jump I can't jump more thank feet. And if I jump exactly k feet in a jump, k is decremented by 1. But if I jump less than k feet, k remains same.

For example, let the height of the rungs from the ground be 1, 6, 7, 11, 13 respectively and k be 5. Now the steps are:

1.      Jumped 1 foot from the ground to the 1st rung (ground to 1). Since I jumped less than k feet, k remains 5.

2.      Jumped 5 feet for the next rung (1 to 6). So, k becomes 4.

3.      Jumped 1 foot for the 3rd rung (6 to 7). So, k remains 4.

4.      Jumped 4 feet for the 4th rung (7 to 11). Thus k becomes 3.

5.      Jumped 2 feet for the 5th rung (11 to 13). And so, k remains 3.

Now you are given the heights of the rungs of the ladder from the ground, you have to find the minimum strength factor k, such that I can reach the top rung.





Input



Input starts with an integer T (≤ 12), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 105) denoting the number of rungs in the ladder. The next line contains n space separated integers, r1, r2 ..., rn (1 ≤ r1 < r2 < ... < rn ≤ 109)denoting the heights of the rungs from the ground.





Output



For each case, print the case number and the minimum possible value of k as described above.





Sample Input



2

5

1 6 7 11 13

4

3 9 10 14





Sample Output



Case 1: 5

Case 2: 6




Source



Problem Setter: Jane Alam Jan


Special Thanks: Sohel Hafiz


//题意：输入一个数字n，接下来一行输入n个数


表示你第 i 步到达的高度，问最小的爬升距离k是多少才能让你爬到杆顶？当然爬升是有规定的，如果你某一步的爬升距离等于k时，k的值就得减1，否则一直是k（某一步的爬升距离不能超过k）。


//思路：


根据题意规定可知，先求出相邻两高度之间的高度差的最大值mm，再用这个最大值mm遍历这n个高度差，看是否满足题意，若不满足，那么k==mm+1,否则k==mm;


Hait：你开始是在0高度处（在这块WA了一次）。。 



#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define N 100010
using namespace std;
int a[N],b[N];
int main()
{
	int t,T=1,n,i,j,k;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		scanf("%d",&a[0]);
		b[0]=a[0];
		int mm=b[0];
		for(i=1;i<n;i++)
		{
			scanf("%d",&a[i]);
			b[i]=a[i]-a[i-1];
			mm=max(mm,b[i]);
		}
		int m=mm;
		int flag=0;
		for(i=1;i<n;i++)
		{
			if(mm==b[i])
				mm--;
			else if(mm<b[i])
			{
				flag=1;
				break;
			}
		}
		printf("Case %d: ",T++);
		if(flag)
			printf("%d\n",m+1);
		else
			printf("%d\n",m);
	}
}