Codeforces Round #506 (Div. 3)B.Creating the Contest(dp)

时间：2018-08-25 14:17:13 阅读：192 评论：0 收藏：0 [点我收藏+]

标签：tis lin ++ cti row tin guarantee codeforce selected

You are given a problemset consisting of $n$

You have to assemble the contest which consists of some problems of the given problemset. In other words, the contest you have to assemble should be a subset of problems (not necessary consecutive) of the given problemset. There is only one condition that should be satisfied: for each problem but the hardest one (the problem with the maximum difficulty) there should be a problem with the difficulty greater than the difficulty of this problem but not greater than twice the difficulty of this problem. In other words, let $a_{i_{1}}, a_{i_{2}}, \dots, a_{i_{p}}$

Among all contests satisfying the condition above you have to assemble one with the maximum number of problems. Your task is to find this number of problems.

Input

The first line of the input contains one integer $n$

The second line of the input contains $n$

Output

Print a single integer — maximum number of problems in the contest satisfying the condition in the problem statement.

Examples

input

Copy

10
1 2 5 6 7 10 21 23 24 49

output

Copy

input

Copy

5
2 10 50 110 250

output

Copy

input

Copy

6
4 7 12 100 150 199

output

Copy

Note

Description of the first example: there are $10$

In the second example all the valid contests consist of $1$

In the third example are two contests consisting of $3$

我怕不是智障数组开小了无限TLE。。。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int n;
 8     int a[200010];
 9     int dp[200010];
10     int result = 0;
11     scanf("%d",&n);
12     dp[0] = 0;
13 
14     for(int i = 1;i<=n;++i)
15     {
16         scanf("%d",&a[i]);
17         dp[i] = 1;
18         if(a[i-1]*2>=a[i])
19                // break;
20                {
21                    if(dp[i]<dp[i-1]+1)
22                     dp[i] = dp[i-1]+1;
23 
24                }
25             if(result<dp[i])
26         result = dp[i];
27     }
28     //for(int i = 2;i<=n;++i)
29     //{
30        // for(int j = i-1;j>=1;--j)
31         //{
32             //if(a[i-1]*2>=a[i])
33                // break;
34            // dp[i] = dp[i-1]+1;
35         //}
36         //result = max(result,dp[i]);
37     //}
38     cout << result<<endl;
39     //cout << *max_element(dp+1,dp+n)<<endl;
40     return 0;
41 }

Codeforces Round #506 (Div. 3)B.Creating the Contest(dp)

标签：tis lin ++ cti row tin guarantee codeforce selected

原文地址：https://www.cnblogs.com/knmxx/p/9533146.html

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