poj1852 Ants

这题考察思维的灵活性，此外这份代码用cin和cout会超时（orz）

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
int a[1000000];
int ans[1000000];
int main()
{
    int t,n,l;
    scanf("%d",&t);
    while(t--)
    {
        int ans1 =-1,ans2 = -1;
        scanf("%d%d",&l,&n);
        for(int i = 1; i <= n; i ++)
        {
            scanf("%d",&a[i]);
        }
        for(int i = 1;i <= n; i ++)
        {
            ans[i] = min(a[i],l-a[i]);
            ans1 = max(ans1,ans[i]);
        }
        for(int i = 1; i <= n; i ++)
        {
            ans[i] = max(a[i],l-a[i]);
            ans2 = max(ans2,ans[i]);
        }
        printf("%d %d\n",ans1,ans2);
    }
    return 0;
}