动态规划——Bubble Sort Graph

Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a₁, a₂, ..., a_n in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode).

procedure bubbleSortGraph()    build a graph G with n vertices and 0 edges    repeat        swapped = false        for i = 1 to n - 1 inclusive do:            if a[i] > a[i + 1] then                add an undirected edge in G between a[i] and a[i + 1]                swap( a[i], a[i + 1] )                swapped = true            end if        end for    until not swapped     /* repeat the algorithm as long as swapped value is true. */ end procedure

For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.

Input

The first line of the input contains an integer n (2 ≤ n ≤ 10⁵). The next line contains n distinct integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n).

Output

Output a single integer — the answer to the problem.

Example

Input

33 1 2

Output

2

Note

Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2].

题意：看了之后才知道是求最长非递减子序列。。。

思路：看数据范围知道是用dp+二分

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int a[100010];int dp[100010];int n;int wtf(int l, int x)   //二分搜索求dp数组中<=x的位置{    int left=1,right=l,mid=(left+right)>>1;    while(left<=right)    {        if(dp[mid]>x)            right=mid-1;        else            if(dp[mid]<x)                left=mid+1;            else                return mid;        mid=(left+right)>>1;    }    return left;}int DP(){    int l=1;    dp[1]=a[0];    for(int i=1; i<n; i++)    {        int k=wtf(l, a[i]);        dp[k]=a[i];        l=max(l, k);    }    return l;}int main(){    while(~scanf("%d", &n))    {        memset(dp, 0, sizeof(dp));        memset(a, 0, sizeof(a));        for(int i=0; i<n; i++)            scanf("%d", &a[i]);        int ans=DP();        printf("%d\n", ans);    }    return 0;}