Creating the Contest
You are given a problemset consisting of nnproblems. The difficulty of the ii-th problem is aiai. It is guaranteed that all difficulties are distinct and are given in the increasing order.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 ai1,ai2,…,aipai1,ai2,…,aip be the difficulties of the selected problems in increasing order. Then for each jj from 11 to p−1p−1 aij+1≤aij⋅2aij+1≤aij⋅2 should hold. It means that the contest consisting of only one problem is always valid.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.InputThe first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of problems in the problemset.The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an(1≤ai≤1091≤ai≤109) — difficulties of the problems. It is guaranteed that difficulties of the problems are distinct and are given in the increasing order.OutputPrint a single integer — maximum number of problems in the contest satisfying the condition in the problem statement.
Examples
Input
10
1 2 5 6 7 10 21 23 24 49
Output
4
Input
5
2 10 50 110 250
Output
1
Input
6
4 7 12 100 150 199
Output
3
NoteDescription of the first example: there are 1010 valid ctests consisting of 11 problem, 1010 alid contests consisting of 22 roblems ([1,2],[5,6],[5,7],[5,10],[6,7],[6,10],[7,10],[21,23],[21,24],[23,24][1,2],[5,6],[5,7],[5,10],[6,7],[6,10],[7,10],[21,23],[21,24],[23,24]), 55 id contests consisting of 33

problems ([5,6,7],[5,6,10],[5,7,10],[6,7,10],[21,23,24][5,6,7],[5,6,10],[5,7,10],[6,7,10],[21,23,24]

) and a single valid contest consisting of 44

problems ([5,6,7,10][5,6,7,10]

).In the second example all the valid contests consist of 11

problem.In the third example are two contests consisting of 33

problems: [4,7,12][4,7,12]

and [100,150,199][100,150,199]

求满足题意的最长子序列的长度
简化LIS：

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
long long a[200050],dp[200050],i,j,n,ans;
int main()
{
   
    while(scanf("%lld",&n)!=EOF)
    {
   
        ans=0;
        for(i=0;i<n;i++)
        {
   
            scanf("%lld",&a[i]);
            dp[i]=1;
            if(i&&a[i]<=2*a[i-1]&&a[i]>a[i-1])
                dp[i]=dp[i-1]+1;
            ans=max(ans,dp[i]);
        }
        cout<<ans<<'\n';
     }
     return0；
 }

暴力扫一遍：

#include <cstdio>
#include <iostream>
using namespace std;
const int N=2e5+40;
int a[N];
int main()
{
   
    int n,i,ans,m;
    while(scanf("%d",&n)!=EOF)
    {
   
        ans=m=1;
        for(i=0;i<n;i++)
        {
   
            scanf("%d",&a[i]);
            if(i&&a[i]<=2*a[i-1])
            {
   
                m++;
                ans=max(ans,m);
            }
            else if(i&&a[i]>2*a[i-1])
                m=1;
        }
        cout<<ans<<'\n';
    }
    return 0;
}

尺取法：对数组保存下标 高效。

#include <bits/stdc++.h>
using namespace std;
const int N=2e5+40;
int a[N];
int main()
{
   
    int ans,n,l;
    while(scanf("%d",&n)!=EOF)
    {
   
        ans=1;
        cin>>a[0];
        l=a[0];
        int j=0;
        for(int i=1;i<n;i++)
        {
   
            cin>>a[i];
            if(a[i]<=2*l)
                ans=max(ans,i-j+1);
            else
                j=i;
            l=a[i];
        }
        cout<<ans<<'\n';
    }
    return 0;
}