2013
11-10

# Ants

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.

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.

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.

2
10 3
2 6 7
214 7
11 12 7 13 176 23 191


4 8
38 207


//* @author: ccQ.SuperSupper
import java.io.*;
import java.util.*;
public class Main {

/**
* @param args
*/
static int max(int a,int b)
{
if(a>b) return a;
return b;
}
static int min(int a,int b)
{
if(a>b) return b;
return a;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int t,n,m,ans1,ans2,i,k;

Scanner cin = new Scanner(System.in);

t = cin.nextInt();
while(t>0)
{
--t;

n = cin.nextInt();
m = cin.nextInt();

ans1 = 0;ans2 = 0;

for(i=0;i< m;++i)
{
k = cin.nextInt();
ans1 = max(ans1,min(k,n-k));
ans2 = max(ans2,max(k,n-k));
}
System.out.println(ans1+" "+ans2);
}

}

}

1. 思路二可以用一个长度为k的队列来实现，入队后判断下队尾元素的next指针是否为空，若为空，则出队指针即为所求。

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4. 第一题是不是可以这样想，生了n孩子的家庭等价于n个家庭各生了一个1个孩子，这样最后男女的比例还是1:1