Codeforces Round #283 (Div. 2) A

解题思路：给出一个递增数列，a1,a2,a3,-----,an.问任意去掉a2到a3之间任意一个数之后，

因为注意到该数列是单调递增的，所以可以先求出原数列相邻两项的差值的最大值max,

得到新的一个数列（比如先去掉a2），该数列相邻两项的差值的最大值为Max1=findmax(max,a3-a1)

再去掉 a3,得到该数列相邻两项的差值的最大值Max2=findmax(max,a4-a2)

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再去掉 an-1,得到该数列相邻两项的差值的最大值Maxn-2=findmax(max,an-an-2)

题目要求的即为 max1,max2,max3,-----,maxn-2的最小值

A.

Minimum Difficulty

time limit per test2 seconds

memory limit per test256

megabytes inputstandard input outputstandard output

Mike is trying rock climbing but he is awful at it.

There are n holds on the wall, i-th hold is at height ai off the ground. Besides, let the sequence ai increase, that is, ai < ai + 1 for all i from 1 to n - 1; we will call such sequence a track. Mike thinks that the track a1, ..., an has difficulty . In other words, difficulty equals the maximum distance between two holds that are adjacent in height.

Today Mike decided to cover the track with holds hanging on heights a1, ..., an. To make the problem harder, Mike decided to remove one hold, that is, remove one element of the sequence (for example, if we take the sequence (1, 2, 3, 4, 5) and remove the third element from it, we obtain the sequence (1, 2, 4, 5)). However, as Mike is awful at climbing, he wants the final difficulty (i.e. the maximum difference of heights between adjacent holds after removing the hold) to be as small as possible among all possible options of removing a hold. The first and last holds must stay at their positions.

Help Mike determine the minimum difficulty of the track after removing one hold.

Input The first line contains a single integer n (3 ≤ n ≤ 100) — the number of holds.

The next line contains n space-separated integers ai (1 ≤ ai ≤ 1000), where ai is the height where the hold number i hangs. The sequence ai is increasing (i.e. each element except for the first one is strictly larger than the previous one).

Output Print a single number — the minimum difficulty of the track after removing a single hold.

Sample test(s) input 3 1 4 6 output 5 input 5 1 2 3 4 5 output 2 input 5 1 2 3 7 8 output 4 Note In the first sample you can remove only the second hold, then the sequence looks like (1, 6), the maximum difference of the neighboring elements equals 5.

In the second test after removing every hold the difficulty equals 2.

In the third test you can obtain sequences (1, 3, 7, 8), (1, 2, 7, 8), (1, 2, 3, 8), for which the difficulty is 4, 5 and 5, respectively. Thus, after removing the second element we obtain the optimal answer — 4.

#include<stdio.h>
int searchmax(int a[],int n)
{
	int i,max;
	max=a[1];
	for(i=2;i<=n;i++)
	{
		if(a[i]>max)
		max=a[i];
	}
	return max;
}
int searchmin(int a[],int n)
{
	int i,min;
	min=a[1];
	for(i=2;i<=n;i++)
	{
		if(a[i]<min)
		min=a[i];
	}
	return min;
}

int findmax(int a,int b)
{
	if(a>b)
	return a;
	else
	return b;
}

int main()
{
	int a[105],d[105],e[105],i,n,s,j,k,max;
	while(scanf("%d",&n)!=EOF)
	{
		k=1;
		max=0;
		for(i=1;i<=n;i++)
		scanf("%d",&a[i]);
		for(i=1;i<n;i++)
		d[k++]=a[i+1]-a[i];
		max=searchmax(d,n-1);
		j=1;
		for(i=2;i<=n-1;i++)
		{
		e[j++]=findmax(max,a[i+1]-a[i-1]);//找到删掉一个数后的数列中，相邻两项的差值的最大值
	    }

		printf("%d\n",searchmin(e,n-2));
	}
}