POJ1852 Ants

Description

An army of ants walk on a horizontal pole of length l cm, each with a constant speed of 1 cm/s. When a walking ant reaches an end of the pole, it immediatelly falls off it. When two ants meet they turn back and start walking in opposite directions. We know the original positions of ants on the pole, unfortunately, we do not know the directions in which the ants are walking. Your task is to compute the earliest and the latest possible times needed for all ants to fall off the pole.

Input

The first line of input contains one integer giving the number of cases that follow. The data for each case start with two integer numbers: the length of the pole (in cm) and n, the number of ants residing on the pole. These two numbers are followed by n integers giving the position of each ant on the pole as the distance measured from the left end of the pole, in no particular order. All input integers are not bigger than 1000000 and they are separated by whitespace.

Output

For each case of input, output two numbers separated by a single space. The first number is the earliest possible time when all ants fall off the pole (if the directions of their walks are chosen appropriately) and the second number is the latest possible such time. 

Sample Input

2
10 3
2 6 7
214 7
11 12 7 13 176 23 191

Sample Output

4 8
38 207

题意：

      n只蚂蚁以每秒1cm的速度在长为L的杆子上爬行，当蚂蚁爬到端点时就会掉下来。两只蚂蚁相遇时，只能返回，对于每只蚂蚁，只知道它距离杆子左边距离，不知道它爬行的方向。计算所有蚂蚁掉下杆子的所需的最短时间和最长时间。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 1000010
int l,n;
int a[maxn];
void solve()
{
	int minT=0;
	for(int i=0;i<n;i++)
	{
		minT=max(minT,min(a[i],l-a[i]));
	}
	int maxT=0;
	for(int i=0;i<n;i++)
	{
		maxT=max(maxT,max(a[i],l-a[i]));
	}
	printf("%d %d\n",minT,maxT);
}
int main()
{
	int t,i;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&l,&n);
		for(i=0;i<n;i++)
		  scanf("%d",&a[i]);
		solve();
	}
	return 0;
}