POJ

Ants

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 18122 Accepted: 7671

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

210 32 6 7214 711 12 7 13 176 23 191

Sample Output

4 838 207 输入N，代表有N组数据，每组数据输入n,m，然后是n个数，分别代表蚂蚁所在位置，蚂蚁每次移动1cm，两只蚂蚁相遇时，前进方向改变，m为棍子长度，问最少多少次移动所有蚂蚁离开木棒，最大多少次移动所有蚂蚁离开木棒。找规律。。
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    int N;    scanf("%d",&N);    while(N--)    {        int n,m;        scanf("%d%d",&m,&n);        int ans1=0,ans2=0;        for(int i=0; i<n; i++)        {            int t;            scanf("%d",&t);            ans1=(ans1>(m-t<t?m-t:t)?ans1:(m-t<t?m-t:t));            ans2=(ans2>(m-t>t?m-t:t)?ans2:(m-t>t?m-t:t));        }        printf("%d %d\n",ans1,ans2);    }    return 0;}