Maximum non-descending sub-sequence

Solution 1 (dichotomy: O (n*log2 (n))):

#include <stdio.h>
#include <iostream>
#define MAXN 10000
using namespace Std;

int A[MAXN], LOW[MAXN];

int main (void)
{
int n;
while (CIN >> N)
{
LOW[1]=A[1];
int now=1;
for (int i=1; i<=n; i++)
CIN >> A[i];
for (int i=2; i<=n; i++)
{
if (A[i]>low[now]) low[++now]=a[i];
Else
{
int Pos=lower_bound (Low, Low+now, A[i]) – low;
Low[pos]=a[i];
}
}
cout << now << Endl;
}
return 0;
}

Solution 2 (Dichotomy queue implementation: O (N*long2 (n))):

#include <stdio.h>
#include <iostream>
#include <string.h>
#define MAXN 1000000
using namespace Std;

int main (void)
{
int n;
while (CIN >> N)
{
int STACK[MAXN];
int top=0;
Stack[0]=-1;
for (int i=1; i<=n; i++)
{
int temp;
CIN >> Temp;
if (Temp>stack[top]) stack[++top]=temp;
Else
{
int Pos=lower_bound (stack, stack+top, temp)-stack;
Stack[pos]=temp;
}
}
cout << top << Endl;
}
return 0;
}

Solution 3 (O (n*n)):

#include <stdio.h>
#include <iostream>
#define MAXN 10000+10
using namespace Std;

int main (void)
{
int n;
while (CIN >> N)
{
int A[MAXN], alen[maxn], maxlen=1;
for (int i=1; i<=n; i++)
Alen[i]=1;
for (int i=1; i<=n; i++)
CIN >> A[i];
for (int i=2; i<=n; i++)
{
int max=0;
for (int j=1; j<=i-1; j + +)
if (A[j]<a[i] && max<alen[j]) max=alen[j];
alen[i]=max+1;
if (Alen[i]>maxlen) maxlen=alen[i];
}
cout << maxlen << Endl;
}
return 0;
}

Method 4 (DP (O (n*n))):

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#define MAXN 100
using namespace Std;

int main (void)
{
int n;
while (CIN >> N)
{
int A[MAXN], LOW[MAXN];
memset (Low, 0, sizeof (low));
for (int i=0; i<n; i++)
CIN >> A[i];
for (int i=0; i<n; i++)
{
for (int j=i+1; j<n; j + +)
{
if (A[j]>a[i])
low[j]++;
}

}
Sort (low, low+n);
cout << low[n-1];
}
return 0;
}

Maximum non-descending sub-sequence