itdef

地址 https://www.papamelon.com/problem/193


解答
根據(jù)題意，螞蟻a和b相遇后各自反向，起始就等于a和b穿過對方繼續(xù)行走。所以沒有任何干擾，就是求每個螞蟻向左或者向右的最大最小時間

#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;

const int N = 1000010;
int arr[N];
int t, n, m;


int main()
{
	cin >> t;
	while (t--) {
		memset(arr, 0, sizeof arr);
		cin >> m >> n;
		for (int i = 0; i < n; i++) {
			cin >> arr[i];
		}

		int a = -1; int b = -1;

		for (int i = 0; i < n; i++) {
			a = max(a, min(m - arr[i], arr[i]));
			b = max(b, max(m - arr[i], arr[i]));
		}

		cout << a << " " << b << endl;

	}

	return 0;
}