lightoj 1425 - The Monkey and the Oiled Bamboo

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Time Limit: 3 second(s)

Memory Limit: 32 MB

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 than k 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 1^st 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 3^rd rung (6 to 7). So, k remains 4.

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

5.      Jumped 2 feet for the 5^th 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 ..., r_n (1 ≤ r₁ < r₂ < ... < r_n ≤ 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	Output for Sample Input
2 5 1 6 7 11 13 4 3 9 10 14	Case 1: 5 Case 2: 6

简单的二分加模拟，怕的就是题目没读懂。

求最小的满足条件的k值。

A[i] - a[i-1] <= k。

相等的话，k--。

显然就是二分。。。

点击打开链接

       :2015
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 100010;
int A[maxn];
int n;
bool Judge(int mid){
	int Max = mid;
	for (int i = 2;i <= n;++i){
		if (A[i] - A[i - 1] > Max) return false;
		else if (A[i] - A[i - 1] == Max) Max--;
	}
	return true;
}
void Solve(){
	if (n == 1) printf("%d\n",A[1]);
	else{
		int ans;
		int high = A[n], low = A[1];
		while(high >= low){
			int mid = (high + low) / 2;
			if (Judge(mid)) {
				high = mid - 1;
				ans = mid;
			}
			else low = mid + 1;
		}
		printf("%d\n",ans);
	}
}
int main()
{	
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int t, icase = 0;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for (int i = 1;i <= n;++i)
			scanf("%d",&A[i]);
		printf("Case %d: ", ++icase);
		Solve();
	}
	return 0;
}