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;}