Ants
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions:24497 | Accepted: 9798 |
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
Source
问题简述;
横杆上有n个蚂蚁，蚂蚁在横杆上运动，两只蚂蚁相遇后会调转方向，蚂蚁的速度都为1，输入中会给两个数据，横杆的长度和蚂蚁数量n，n个蚂蚁的初始位置……，要求输出蚂蚁从横杆上掉下的最早时间和最晚时间（自己寻找合适的方向）
分析：
问题的关键是想到可以把蚂蚁“相遇后掉头”的问题等价为“相遇后方向不变 互不干扰”
还要想到max，min函数来简化程序
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int main(){
int t;
scanf("%d",&t);
while(t--){
int l,n;
int minn=0;
int maxn=0;
scanf("%d%d",&l,&n);
int val;
for(int i=0;i<n;i++){
scanf("%d",&val);
minn=max(minn,min(val,l-val));
maxn=max(maxn,max(val,l-val));
}
printf("%d %d\n",minn,maxn);
}
return 0;
}