首页 > 代码库 > codeforces A. TL 题解

codeforces A. TL 题解

2024-07-02 16:51:07   231人阅读

Valera wanted to prepare a Codesecrof round. He‘s already got one problem and he wants to set a time limit (TL) on it.

Valera has written n correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote m wrong solutions and for each wrong solution he knows its running time (in seconds).

Let‘s suppose that Valera will set v seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most v seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, a seconds, an inequality 2a?≤?v holds.

As a result, Valera decided to set v seconds TL, that the following conditions are met:

  1. v is a positive integer;
  2. all correct solutions pass the system testing;
  3. at least one correct solution passes the system testing with some "extra" time;
  4. all wrong solutions do not pass the system testing;
  5. value v is minimum among all TLs, for which points 1, 2, 3, 4 hold.

Help Valera and find the most suitable TL or else state that such TL doesn‘t exist.

Input

The first line contains two integers n, m (1?≤?n,?m?≤?100). The second line contains n space-separated positive integers a1,?a2,?...,?an (1?≤?ai?≤?100) — the running time of each of the n correct solutions in seconds. The third line contains m space-separated positive integers b1,?b2,?...,?bm (1?≤?bi?≤?100) — the running time of each of mwrong solutions in seconds.

Output

If there is a valid TL value, print it. Otherwise, print -1.

Sample test(s)
input
3 6
4 5 2
8 9 6 10 7 11
output
5

这句话难读懂：

We can also say that a solution passes the system testing with some "extra" time if for its running time, a seconds, an inequality 2a?≤?v holds.

就是说：如果一个AC的解决方案的运行时间是a，满足2*a <= v，那么就说这个运行时间通过而且有"extra"时间(a)


void TLjudge()
{
	int n, m, minVal = 1<<30, maxVal = 1<<31, a = 0;
	cin>>n>>m;
	while (n--)
	{
		cin>>a;
		if (a < minVal) minVal = a;
		if (a > maxVal) maxVal = a;
	}
	int ans = max(minVal*2, maxVal);
	bool ok = true;
	while (m--)
	{
		cin>>a;
		if (a <= ans)
		{
			ok = false;
			break;
		}
	}
	if (ok) cout<<ans;
	else cout<<-1;
};


联系
我们