poj 1852 Ants (思维)

Ants

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 13918 Accepted: 6072

Description

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.

Input

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.

Output

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. 

Sample Input

210 32 6 7214 711 12 7 13 176 23 191

Sample Output

4 8
38 207
嗯。看了别人的题解，才明白那么简单，最少用多少时间能全掉下去 就向较近极点走的离端点最远的点；最多用多少时间全掉下去 就向较远的端点走 最远的点！蚂蚁相遇可认为直接穿过继续走不掉头！（像动量定理得质量相等的速度交换）
代码：
#include <iostream>#include <cstdio>using namespace std;int a[1000010];int min(int x,int y)<span style="font-family: 'Courier New', Courier, monospace;">这里是第一次这样用 以后可以多加练习</span>{   return x<y?x:y;}int max(int x,int y){    return x>y?x:y;}int main(){    int t,l,n;    int minn,maxx;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&l,&n);        minn=0;        maxx=0;        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);            minn=max(minn,min(a[i],l-a[i]));//这里也是第一次这样用 以后可以多加练习            maxx=max(maxx,max(a[i],l-a[i]));        }        printf("%d %d\n",minn,maxx);    }    return 0;}