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Description

    On the PE，the teacher wants to choose some of n students to play games. Teacher asks n students stand in a line randomly(obviously，they have different height), then teacher tells someone to leave the queue（relative position is not change）to make the left students keep a special queue：

    Assuming that the left students’ numbers are 1, 2,……, m from left to right and their height are T1,T2,……,Tm. Then they satisfy T1<T2<......<Ti, Ti>Ti+1>……>Tm (1<=i<=m).     Giving you the height of n students (from left to right), please calculate how many students left at most if you want to keep such a special queue.  

Input

    The first line contains an integer n (2<=n<=1000) represents the number of students.

    The second line contains n integers Ti(150<=Ti<=200) separated by spaces, represent the height(cm) of student i.

Output

    One integer represents the number of the left students.

Sample Input

8186 180 150 183 199 130 190 180

Sample Output

5

n个学生，求满足身高左递增右递减的最长序列 可以没有递增部分或没有递减部分

正序 反序分别求一遍LIS，然后遍历每个点，正序+反序-1 即为结果 -1是因为该点会重复一次

#include 
   
    using namespace std;int dp1[1005],dp2[1005],a[1005],b[1005];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            b[n-i+1]=a[i];        }        memset(dp1,0,sizeof(dp1));        memset(dp2,0,sizeof(dp2));        dp1[1]=1;        dp2[1]=1;        for(int i=2;i<=n;i++)        {            dp1[i]=1;            dp2[i]=1;            for(int j=1;j
   

 

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