Codeforces Round #FF (Div. 2)C. DZY Loves Sequences



C. DZY Loves Sequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

DZY has a sequence a, consisting of n integers.

We'll call a sequence a_i, a_i + 1, ..., a_j(1 ≤ i ≤ j ≤ n) a subsegment of the sequencea. The value (j - i + 1) denotes the length of the subsegment.

Your task is to find the longest subsegment of a, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.

You only need to output the length of the subsegment you find.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵). The next line containsn integersa₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Output

In a single line print the answer to the problem — the maximum length of the required subsegment.

Sample test(s)

Input

67 2 3 1 5 6

Output

5

Note

You can choose subsegment a₂, a₃, a₄, a₅, a₆ and change its 3rd element (that is a₄) to 4.

考察最长连续上升子序列的一道题，dp经典题的变形题

dp[i][0]表示以i结尾的最长连续子列，dp[i][1]表示以i开始的最长子列

最后枚举i的改变，注意i的改变有三种情况，1使左半部分取最大值，2使右半部分取最大值，3使左右连续起来取最大值

</pre><pre class="cpp" name="code">#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=100000+10;int dp[maxn][2];int a[maxn];int main(){int n;while(scanf("%d",&n)!=EOF){for(int i=1;i<=n;i++)scanf("%d",&a[i]);dp[1][0]=1;for(int i=2;i<=n;i++){if(a[i]>a[i-1]) dp[i][0]=dp[i-1][0]+1;elsedp[i][0]=1;}dp[n][1]=1;for(int i=n-1;i>=1;i--){if(a[i]<a[i+1])dp[i][1]=dp[i+1][1]+1;elsedp[i][1]=1;}int ans=1;int dd;for(int i=2;i<=n-1;i++){if((a[i+1]-a[i-1])>1){dd=dp[i-1][0]+dp[i+1][1]+1;    ans=max(ans,dd);}else{ans=max(ans,dp[i][0]+1);}}for(int i=1;i<=n-1;i++){ans=max(ans,dp[i+1][1]+1);}for(int i=2;i<=n;i++)ans=max(ans,dp[i-1][0]+1);//for(int i=1;i<=n;i++)//cout<<dp[i][1]<<endl<<dp[i][0]<<endl;cout<<ans<<endl;}return 0;}