Predicated upon the dual conditions of your nativity occurring in the chronological bounds of 2008, concurrently coupled with your geographical domicile within

Question

Predicated upon the dual conditions of your nativity occurring in the chronological bounds of 2008, concurrently coupled with your geographical domicile within the sovereign territory of Vietnam, I harbor the steadfast conviction that you are endowed with the requisite cognitive faculties to successfully unravel this particular conundrum (LOL hahahaha)
Longest Increasing Subsequence
Problem Statement
You are given an array of N integers: `A[1], A[2], ..., A[N]`
A subsequence is considered strictly increasing if it can be formed by selecting elements such that `A[i_1] < A[i_2] < ... < A[i_k]`, where their original indices satisfy `i_1 < i_2 < ... < i_k`.
Task:
Determine the length of the longest strictly increasing subsequence of the given array.
Input Format
* The first line contains a single integer `N (1 <= N <= 30000)`, representing the size of the array.
* The second line contains N space-separated integers `A[1], A[2], ..., A[N] (0 <= A[i] <= 1000000)`.
Output Format
* Print a single integer representing the maximum length of the increasing subsequence.
Example:
Input:
6
1 2 5 4 6 2
Output:
4
The scope of admissible development tools is strictly restricted to the following programming languages: C, Objective-C, Haskell, OCaml, and Pascal.

haolac · Answer

Pascal :
uses math;
const
    k = 30000;
var
    n, i, l, p: longint;
    a,t: array[1..k] of longint;
function binarysearch(x, y, z: longint): longint;
var
    m, s: longint;
begin
    s:=z + 1;
    while y=x then
        begin
            s:=m;
            z:=m-1;
        end
        else
            y:=m+1;
    end;
    exit(s);
end;

begin
    readln(n);
    for i:=1 to n do
        read(a[i]);
    l:=0; 
    for i:=1 to n do
    begin
        p:=binarysearch(a[i],1,l);
        t[p]:=a[i];
        if p>l then
            l:=p;
    end;
    writeln(l);
end.
#Haolac

tuandung16 · Answer

`\color{black}{#Jadess౨ৎ}`
#include 
using namespace std;

int main() {
    int n;
    cin >> n;
    vector a(n), d;
    for (int i = 0; i > a[i];
        auto it = lower_bound(d.begin(), d.end(), a[i]);
        if (it == d.end()) d.push_back(a[i]);
        else *it = a[i];
    }
    cout 
 `\color{#1AD5F7}{꧁⋆⟡T}\color{#1AD5F7}{D}\color{#4DA6E6}{u}\color{#668EDD}{n}\color{#8077D5}{g}\color{#B28DFF}{\ }\color{#995FCD}{G}\color{#EA2F90}{M⟡⋆꧂}`

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