(POJ) 1852--ants (ANT)

Describe
The ants ' armies walk on the horizontal poles of the length LCM, each with a constant speed of 1cm/s. When a walking ant reaches one end of the pole, it drops immediately. When two ants meet, they return and start moving in the opposite direction. We know the original position of the ant on the pole, unfortunately, we don't know the direction of the ant walking. Your task is to calculate the shortest and longest possible time for all ants to get out of the pole.

Input
The first line of input contains an integer that gives the number of cases. The data for each case starts with two integers: the length of the pole (in cm), and n is the number of ants on the pole. These two numbers follow the n integers, giving the position of each ant on the pole as the distance measured from the left side of the bar, without a specific order. All input integers are not greater than 1000000, and they are separated by spaces.

Output
For each case input, the output is two digits separated by a space. The first number is the earliest possible time for all ants to leave the pole (if their direction of walk is chosen appropriately), the second number is the longest 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

This topic can ignore the collision of ants, will each ant's Movement shortest time and the longest time to seek out, respectively take inside the maximum value.

1#include <iostream>2#include <cstdio>3#include <algorithm>4 5 using namespacestd;6 7 intMain ()8 {9     intt,x,i;Tenscanf"%d",&t); One      while(t--) A     { -         intL,n,max_time=0, min_time=0; -scanf"%d%d",&l,&n); the          for(i=1; i<=n;i++) -         { -scanf"%d",&x); -Max_time=max (Max_time,max (x,l-x)); +Min_time=max (Min_time,min (x,l-x)); -         } +printf"%d%d\n", min_time,max_time); A     } at     return 0; -}

Answer

(POJ) 1852--ants (ANT)