C. Multi-judge Solving 贪心+模拟

最新推荐文章于 2019-08-07 10:11:00 发布

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最新推荐文章于 2019-08-07 10:11:00 发布

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分类专栏：暑期集训刷题算法复习（新手）

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本文链接：https://blog.youkuaiyun.com/FrankAx/article/details/76975078

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本文介绍了一种解决特定类型编程挑战的有效策略。通过合理的排序和条件检查，该方法能够最小化为完成一系列任务所需额外解决的问题数量。文章提供了一个实际的C++实现案例。

Makes solves problems on Decoforces and lots of other different online judges. Each problem is denoted by its difficulty — a positive integer number. Difficulties are measured the same across all the judges (the problem with difficulty d on Decoforces is as hard as the problem with difficulty d on any other judge).

Makes has chosen n problems to solve on Decoforces with difficulties a₁, a₂, ..., a_n. He can solve these problems in arbitrary order. Though he can solve problem i with difficulty a_i only if he had already solved some problem with difficulty $\text{[math]}$ (no matter on what online judge was it).

Before starting this chosen list of problems, Makes has already solved problems with maximum difficulty k.

With given conditions it's easy to see that Makes sometimes can't solve all the chosen problems, no matter what order he chooses. So he wants to solve some problems on other judges to finish solving problems from his list.

For every positive integer y there exist some problem with difficulty y on at least one judge besides Decoforces.

Makes can solve problems on any judge at any time, it isn't necessary to do problems from the chosen list one right after another.

Makes doesn't have too much free time, so he asked you to calculate the minimum number of problems he should solve on other judges in order to solve all the chosen problems from Decoforces.

Input

The first line contains two integer numbers n, k (1 ≤ n ≤ 10³, 1 ≤ k ≤ 10⁹).

The second line contains n space-separated integer numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Output

Print minimum number of problems Makes should solve on other judges in order to solve all chosen problems on Decoforces.

Examples

Input

3 3
2 1 9

Output

Input

4 20
10 3 6 3

Output

Note

In the first example Makes at first solves problems 1 and 2. Then in order to solve the problem with difficulty 9, he should solve problem with difficulty no less than 5. The only available are difficulties 5 and 6 on some other judge. Solving any of these will give Makes opportunity to solve problem 3.

In the second example he can solve every problem right from the start.

题意：要求做完给出的所有难度的题，当给出的题的难度大于K的2倍时，必须从其他OJ做题以此更新K值让2*k大于当前题目才能解题成功，问做完需要额外做几题。

思路：从小到大排序，一道道做，不能做就一直乘2；更新K，能做就比较K和a[i]的大小，更新k

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int AX = 1e3+666;
LL a[AX];
LL n ,k ;
int main(){

	while( cin>>n>>k ){
		for( int i = 1 ; i <= n; i++ ){
			cin>>a[i];
		}
		sort( a + 1 , a + n + 1 );
		int ans = 0;
		for( int i = 1 ; i <= n; i++ ){
			if( a[i] > 2*k ){
				k *= 2 ;
				ans++;
				i--;        //这里wa了，一定要多次判断，开始没太明白题意。。
			}else{
				k = max(a[i],k);
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}