poj 1852Ants问题

Ants

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12697		Accepted: 5557

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

#include 
#include
using namespace std;
void solve(int n, int L, int x[])
{
	int minT = 0;
	int maxT = 0;
	for (int i = 0;i < n;i++) 
	{
		minT = max(minT, min(x[i], L - x[i]));
		maxT = max(maxT, max(x[i], L - x[i]));
	}
	cout << minT << ' ' << maxT << endl;//注意这里题中要求的单空格。
}
int main()
{
	int L, n,m;
	int x[100000];//笔者一开始把数组大小调小了竟然也不通过···
	cin >> m;
//两层循环实现读入两组数据
	for (int j = 1;j <= m;j++)
	{
		cin >> L >> n;
		for (int i = 0;i < n;i++)
		{
			cin >> x[i];
		}
		solve(n, L, x);
	}
	return 0;
}

总结：

这个题做了不是很久，但是真的好多细节要注意。

1.数组的大小

2.题目要求的输出格式

3.注意到最少的时间就是各个蚂蚁向着离自己最近的地方跑，因为各个蚂蚁的速度相等。