POJ 1852 Ants

Source: Internet
Author: User
Tags cas integer numbers
Ants
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 12661 Accepted: 5545

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

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

Sample Output

4 8
38 207

Source

A sad question. ~~,MAXN and Minn are just beginning to define global variables, several times, and later found not all initialized to 0 ... ( ̄o ̄). Z Z ( ̄o ̄). Z Z

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int main ()
{
    int n,cas,l,temp;
    Cin>>cas;
    for (int i=1;i< =cas;i++)
    {
         cin>>l>>n;
         int minn=0,maxn=0;
         for (int i=1;i<=n;i++)
         {
              cin>>temp;
              Minn=max (Minn,min (temp,l-temp));
              Maxn=max (Maxn,max (temp,l-temp));
         }
         cout<<minn<< "" <<maxn<<endl;
   }
    return 0;
}

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