poj 1852 Ants

B - Problem B
Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u
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Status
 
Practice
 
POJ 1852
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
2
10 3
2 6 7
214 7
11 12 7 13 176 23 191
Sample Output
4 8

38 207

最短时间自己想，，，，，

最长时间是蚂蚁开始想中间跑，，，当蚂蚁相遇时，可以把蚂蚁对换，，继续向一边走，，（就是不管别的蚂蚁，每一蚂蚁向距它远的方向走，，，，）

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longint cc,n;int shu[1001005];int main(){int t;scanf("%d",&t);while (t--){scanf("%d%d",&cc,&n);for (int i=0;i<n;i++)scanf("%d",&shu[i]);int zz=0,dd=0,lp;for (int i=0;i<n;i++){lp=cc-shu[i];zz=max(zz,max(lp,shu[i]));dd=max(dd,min(lp,shu[i]));}printf("%d %d\n",dd,zz);}return 0;}