poj -1852 ants (思维题）

n只蚂蚁以每秒1cm的速度在长为Lcm的杆子上爬行，当蚂蚁爬到杆子的端点时就会掉落，由于杆子太细，两只蚂蚁相遇时，他们不能交错通过，只能各自反向爬回去，对于每只蚂蚁，我们知道它距离杆子左端的距离为x，但不知道它当前的朝向，请计算所有蚂蚁落下杆子所需的最短时间很最长时间。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)

#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("a.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);
#define N 1000005
using namespace std;

int f[N];
int main()
{
   // freopen("a.txt","r",stdin);
    int t,l,n,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&l,&n);
        for(i=0;i<n;i++)
            scanf("%d",&f[i]);
        int minn=0;
        for(i=0;i<n;i++)
        {
            minn=max(minn,min(f[i],l-f[i]));
        }
        int maxn=0;
        for(i=0;i<n;i++)
        {
            maxn=max(maxn,max(f[i],l-f[i]));
        }
        printf("%d %d\n",minn,maxn);
    }
    return 0;
}