poj 1852 Ants

Ants

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 33420		Accepted: 12441

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 8
38 207

对于最短时间，所有蚂蚁都朝向较近的端点走会比较好，这种情况下不会发生两只蚂蚁相遇的情况，而且也不可能在比此种情况更短的时间内走到杆子的端点。

考虑最长时间，假如两只蚂蚁相遇，两只蚂蚁都朝相反的方向走，而我们可以认为这两只蚂蚁保持原样交错而行。可以认为每只蚂蚁都是独立运动的。

#include <algorithm>
#include <cstdio>
using namespace std;

int main(){
    int test, l, n;
    scanf("%d", &test);
    while(test--){
        int least = 0, most = 0;
        scanf("%d %d", &l, &n);
        int x;
        for(int i = 0; i < n; i++){
            scanf("%d", &x);
            least = max(least, min(x, l - x));
            most = max(most, max(x, l - x));
        }
        printf("%d %d\n", least, most);
    }
    return 0;
}