poj1852 Ants ——想法题、水题

求最短时间和最长时间。

当两个蚂蚁相遇的时候，可以看做两个蚂蚁穿过，对结果没有影响。O（N）的复杂度

c++版：

#include <cstdio>
#define min(a, b) (a) < (b) ? (a) : (b)
#define max(a, b) (a) > (b) ? (a) : (b)
int main(void)
{
  #ifndef ONLINE_JUDGE
  freopen("in.txt", "r", stdin);
  #endif // ONLINE_JUDGE
  int Case;
  scanf("%d", &Case);
  int n, L, Min, Max;
  while (Case--) {
    Min = -1; Max = -1;
    scanf("%d%d", &L, &n);
    int tmp;
    while (n--) {
      scanf("%d", &tmp);
      int t1 = min(tmp, L - tmp);
      int t2 = max(tmp, L - tmp);
      if (Min < t1)
        Min = t1;
      if (Max < t2)
        Max = t2;
    }
    printf("%d %d\n", Min, Max);
  }
  return 0;
}

Java版：

import java.util.*;
public class ants {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner input = new Scanner(System.in);
        int Case;
        Case = input.nextInt();
        while (Case-- != 0) {
            int L;
            int n;
            L = input.nextInt();
            n = input.nextInt();
            int Max, Min;
            Max = Min = 0;
            while (n-- != 0) {
                int tmp;
                tmp = input.nextInt();
                int Left = tmp, Right = L - tmp;
                if (Left < Right) {
                    Min = (Min < Left ? Left : Min);
                    Max = (Max < Right ? Right : Max);
                }
                else {
                    Min = (Min < Right ? Right : Min);
                    Max = (Max < Left ? Left : Max);
                }
            }
            System.out.println(Min + " " + Max);
        }
    }

}

其实Java写起来也不错的。