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:
Help Valera and find the most suitable TL or else state that such TL doesn‘t exist.
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.
If there is a valid TL value, print it. Otherwise, print -1.
3 6 4 5 2 8 9 6 10 7 11
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; };
codeforces A. TL 题解,布布扣,bubuko.com
原文:http://blog.csdn.net/kenden23/article/details/24873129