A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, the sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e.g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences of this sequence are of length 4, e.g., (1, 3, 5, 8).
Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.
Input
The first line of input contains the length of sequence N (1 <= N <= 1000). The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces.
Output
Output must contain a single integer - the length of the longest ordered subsequence of the given sequence.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Sample Input
1
7
1 7 3 5 9 4 8
Sample Output
4
1：DP
2：DP+队列优化
3：BIT（我自己想的）
虽然dp很粗暴霸气，但突然疑问到，能不能用BIT来实现呢，好像可以试一试。
于是乎：
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<memory.h>
using namespace std;
int a[10001],ans,n,x,temp,Max;
int _S()
{
char c=getchar();
while(c>'9'||c<'0') c=getchar();
int s=0;
while(c<='9'&&c>='0'){
s=s*10+c-'0';
c=getchar();
}
return s;
}
void _add(int v,int num)
{
while(v<=10000){
if(a[v]<num) a[v]=num;
v+=(-v&v);
}
}
int _Max(int v)
{
int tans=0;
while(v>0){
if(a[v]>tans) tans=a[v];
v-=(-v&v);
}
return tans;
}
void _init()
{
Max=0;
n=_S();
for(int i=1;i<=n;i++){
x=_S();
temp=_Max(x);
if(temp>Max) Max=temp;
_add(x+1,temp+1);
}
printf("%d
",Max+1);
}
int main()
{
int T;
T=_S();
while(T--){
memset(a,0,sizeof(a));
_init();
if(T)cout<<endl;
}
return 0;
}