| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 17694 | Accepted: 7520 |
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
4 838 207
/*
Ants（ POJ No.1852）
n 只蚂蚁以每秒 1cm 的速度在长为 Lcm 的竿子上爬行。当蚂蚁爬到竿子的端点时就会掉落。
由于竿子太细，两只蚂蚁相遇时，它们不能交错通过，只能各自反向爬回去。对于每只蚂蚁，
我们知道它距离竿子左端的距离 xi，但不知道它当前的朝向。请计算所有蚂蚁落下竿子所需
的最短时间和最长时间。
限制条件
1 ≤ L ≤ 1e+6
1 ≤ n ≤ 1e+6
0 ≤ xi ≤ L
*/
//注意是所有蚂蚁落下竿子所需的最短时间和最长时间，把题目理解错容易gg
#include<cstdio>
#include<iostream>
using namespace std;
const int maxn=1e+6+10;
int main(){
int T,L,n,x;
scanf("%d",&T);
while(T--){
int mi=0,mx=0;
scanf("%d%d",&L,&n);
for(int i=0;i<n;i++)
{
scanf("%d",&x);
mi=max(mi,min(x,L-x));
mx=max(mx,max(x,L-x));
}
printf("%d %d\n",mi,mx);
}
return 0;
}