（SPOJ

Time limit1948 msMemory limit1572864 kBCode length Limit50000 B

Given a list of numbers A output the length of the longest increasing subsequence. An increasing subsequence is defined as a set {i0 , i1 , i2 , i3 , … , ik} such that 0 <= i0 < i1 < i2 < i3 < … < ik < N and A[ i0 ] < A[ i1 ] < A[ i2 ] < … < A[ ik ]. A longest increasing subsequence is a subsequence with the maximum k (length).

i.e. in the list {33 , 11 , 22 , 44}

the subsequence {33 , 44} and {11} are increasing subsequences while {11 , 22 , 44} is the longest increasing subsequence.

Input

First line contain one number N (1 <= N <= 10) the length of the list A.

Second line contains N numbers (1 <= each number <= 20), the numbers in the list A separated by spaces.

Output

One line containing the lenght of the longest increasing subsequence in A.

Example

Input:
5
1 4 2 4 3
Output:
3
题意：求最长递增序列的长度
分析：可以利用记忆化取最大的来实现 如下解法1
解法1： 时间复杂度O(n^2)

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N=15;#define mem(a,n) memset(a,n,sizeof(a))int dp[N],a[N];int n;int dfs(int sta){    int& ret=dp[sta+1];    if(ret!=-1) return ret;    ret=0;    for(int i=sta+1; i<n; i++)    {        if(sta==-1||a[i]>a[sta])        {            ret=max(ret,dfs(i)+1);            //printf("%d  %d\n",i,ret);        }    }    return ret;}int main(){    while(~scanf("%d",&n))    {        mem(dp,-1);        for(int i=0; i<n; i++)            scanf("%d",&a[i]);        //for(int i=0;i<n;i++)        //  printf("%d\n",dp[i]);        printf("%d\n",dfs(-1));    }    return 0;}

解法2： 内部实现过程见：https://www.felix021.com/blog/read.php?1587
时间复杂度O(nlogn)

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define mem(a,n) memset(a,n,sizeof(a))const int INF=0x3f3f3f3f;const int N=105;int a[N],dp[N];int n;void solve(){    for(int i=0;i<n;i++)        *lower_bound(dp,dp+n,a[i])=a[i];    printf("%d\n",lower_bound(dp,dp+n,INF)-dp);}int main(){    while(~scanf("%d",&n))    {        mem(dp,INF);        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        solve();    }    return 0;}