Simple problem of POJ 1852 Ants thinking problem

Ants

Description

An army of ants walk in 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 the ants meet they turn back and the start walking in opposite directions. We know the original positions of ants on the pole, unfortunately, we don't know the directions in which the ants is Wal King. 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 both integer numbers:the length of the pole (in cm) and N, the number of ants residing o n the pole. These numbers is followed by n integers giving the position of all ant on the pole as the distance measured from the Left end of the pole, in no particular order. All input integers is not bigger than 1000000 and they is separated by whitespace.

Output

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

Sample Input

210 32 6 7214 711 12 7 13 176 23 191

Sample Output

4 838 207

1#include <cstdio>2#include <cmath>3 using namespacestd;4 5 intMain ()6 {7     inttest;8 9scanf"%d",&test);Ten  One      while(test--) A     { -         intn,l; -scanf"%d%d",&l,&n); the  -         intleft=L; -         intright=0; -         intmid=l+1; +         DoubleDis_mid= (DoubleL2; -  +         intcur; A  at          for(intI=1; i<=n;i++) -         { -scanf"%d",&cur); -  -             if(cur<Left ) -left=cur; in             if(cur>Right ) -right=cur; to             if(ABS (Cur-dis_mid) <abs (mid-dis_mid)) +Mid=cur; -         } the  *         intMin= (l-mid) <mid? (L-mid): mid; $         intmax= (l-left) >left? (L-Left ): left;Panax Notoginseng         if(right>max) -max=Right ; the         if(l-right>max) +max=l-Right ; A  theprintf"%d%d\n", Min,max); +  -     } $     return 0; $}

View Code

Simple problem of POJ 1852 Ants thinking problem