From 0b536af703acd2c796f4e472eed19565751b15a3 Mon Sep 17 00:00:00 2001
From: 33Arsenic75 <122697564+33Arsenic75@users.noreply.github.com>
Date: Fri, 28 Jun 2024 01:11:56 +0530
Subject: [PATCH] .

---
 problem-of-the-day/day26/B_I_Hate_1111.cpp  | 48 ++++++++++++++
 problem-of-the-day/day26/Good Sequences.cpp | 73 +++++++++++++++++++++
 problem-of-the-day/day26/Kth factor.cpp     | 19 ++++++
 problem-of-the-day/day26/LCM Challenge.cpp  | 45 +++++++++++++
 4 files changed, 185 insertions(+)
 create mode 100644 problem-of-the-day/day26/B_I_Hate_1111.cpp
 create mode 100644 problem-of-the-day/day26/Good Sequences.cpp
 create mode 100644 problem-of-the-day/day26/Kth factor.cpp
 create mode 100644 problem-of-the-day/day26/LCM Challenge.cpp

diff --git a/problem-of-the-day/day26/B_I_Hate_1111.cpp b/problem-of-the-day/day26/B_I_Hate_1111.cpp
new file mode 100644
index 0000000..778312e
--- /dev/null
+++ b/problem-of-the-day/day26/B_I_Hate_1111.cpp
@@ -0,0 +1,48 @@
+/*The main idea is that we can split the numbers into the two halves, the big half and small half,
+ we can place the bigger half at the odd positions and the smaller half at the even positions.
+
+This works because the smallest big number is larger than the biggest small number.
+ Hence, the mean of any two small numbers is smaller than any big number, 
+and the mean of any two big numbers is bigger than any small number.*/
+#include <bits/stdc++.h>
+using namespace std;
+#define ll long long
+#define ii pair<ll,ll>
+#define iii pair<ii,ll>
+#define fi first
+#define se second
+#define endl '\n'
+#define debug(x) cout << #x << " is " << x << endl
+
+#define pub push_back
+#define pob pop_back
+#define puf push_front
+#define pof pop_front
+#define lb lower_bound
+#define ub upper_bound
+
+#define rep(x,start,end) for(auto x=(start);x!=(end);((start)<(end)?x++:x--))
+#define all(x) (x).begin(),(x).end()
+#define sz(x) (int)(x).size()
+
+int n;
+int arr[55];
+
+int main(){
+    ios::sync_with_stdio(0);
+    cin.tie(0);
+    cout.tie(0);
+    cin.exceptions(ios::badbit | ios::failbit);
+    
+    int t;
+    cin >> t;
+    while (t--){
+        cin >> n;
+        rep(x, 0, 2 * n) cin >> arr[x];
+        
+        sort(arr, arr + 2 * n);
+        
+        rep(x, 0, n) cout << arr[x] << " " << arr[x + n] << " ";
+        cout << endl;
+    }
+}
diff --git a/problem-of-the-day/day26/Good Sequences.cpp b/problem-of-the-day/day26/Good Sequences.cpp
new file mode 100644
index 0000000..0468c19
--- /dev/null
+++ b/problem-of-the-day/day26/Good Sequences.cpp	
@@ -0,0 +1,73 @@
+/*The main idea is DP. Let's define dp[x] as the maximal value of the length of the good sequence whose last element is x, 
+and define d[i] as the (maximal value of dp[x] where x is divisible by i).
+
+You should calculate dp[x] in the increasing order of x.
+The value of dp[x] is (maximal value of d[i] where i is a divisor of x) + 1. 
+After you calculate dp[x], for each divisor i of x, you should update d[i] too.
+
+This algorithm works in O(nlogn) because the sum of the number of the divisor from 1 to n is O(nlogn).
+
+Note that there is a corner case. When the set is {1}, you should output 1.
+*/
+
+
+#include <bits/stdc++.h>
+using namespace std;
+#define ll long long
+#define ld long double
+#define vl vector<ll>
+#define pb push_back
+#define pop pop_back
+#define vi vector<int>
+#define vb vector<bool>
+#define rep(i,n) for (ll i = 0; i < (ll)(n); i++)
+#define all(v) v.begin(), v.end()
+
+vl divisor(ll n){
+    vl div(0);
+    set<ll> s;
+    for (ll i = 2; i * i <= n ; i++)
+    {
+        if(s.find(i) != s.end()){
+            break;
+        }
+        if(n % i == 0){
+            s.insert(i);
+            s.insert(n / i);
+            div.pb(i);
+            div.pb(n / i);
+        }
+    }
+    div.pb(n);
+    return div;
+}
+int main(){
+    ios_base::sync_with_stdio(0), cin.tie(0);
+    ll n;
+    cin>>n;
+    vl a(n);
+    for (int i = 0; i < n; i++)
+    {
+        cin>>a[i];
+    }
+    if(n == 1){
+        cout<<1<<endl;
+    }else{
+        vl dp(n);
+        vl d(a[n - 1] + 1);
+        for(int i = 0; i < n; i++){
+            vl div = divisor(a[i]);
+            ll maxi = 0; 
+            for (auto it : div)
+            {
+                maxi = max(d[it], maxi);
+            }
+            dp[i] = max(dp[i], maxi + 1);
+            for (auto it : div)
+            {
+                d[it] = dp[i];
+            }
+        }
+        cout<<*max_element(all(dp))<<endl;
+    }
+}
diff --git a/problem-of-the-day/day26/Kth factor.cpp b/problem-of-the-day/day26/Kth factor.cpp
new file mode 100644
index 0000000..33759c4
--- /dev/null
+++ b/problem-of-the-day/day26/Kth factor.cpp	
@@ -0,0 +1,19 @@
+class Solution {
+ public:
+  int kthFactor(int n, int k) {
+    // If i is a divisor of n, then n / i is also a divisor of n. So, we can
+    // find all the divisors of n by processing the numbers <= sqrt(n).
+    int factor = 1;
+    int i = 0;  // the i-th factor
+
+    for (; factor * factor < n; ++factor)
+      if (n % factor == 0 && ++i == k)
+        return factor;
+
+    for (factor = n / factor; factor >= 1; --factor)
+      if (n % factor == 0 && ++i == k)
+        return n / factor;
+
+    return -1;
+  }
+};
\ No newline at end of file
diff --git a/problem-of-the-day/day26/LCM Challenge.cpp b/problem-of-the-day/day26/LCM Challenge.cpp
new file mode 100644
index 0000000..f04d3b6
--- /dev/null
+++ b/problem-of-the-day/day26/LCM Challenge.cpp	
@@ -0,0 +1,45 @@
+/*It is a simple problem, but many competitors used some wrong guesses and failed. First of all, we should check
+if n is at most 3 and then we can simply output 1,2,6.
+
+Now there are two cases: When n is odd, the answer is obviously n(n-1)(n-2). When n is even,
+we can still get at least (n-1)(n-2)(n-3), so these three numbers in the optimal answer would not be very small compared to n.
+So we can just iterate every 3 number triple in [n-50,n] and update the answer.*/
+
+#include <iostream>
+#include <vector>
+#include <set>
+#include <map>
+#include <cmath>
+#include <iomanip>
+#include <queue>
+#include <algorithm>
+using namespace std;
+typedef long long ll;
+
+ll lcm(ll a ,ll b){
+    return a*b / (__gcd(a,b));
+}
+
+int main()
+{
+    long long n;
+    cin >> n;
+    long long ans = 0;
+    if (n % 2)
+    {
+        ans = n * (n - 1) * (n - 2);
+        if (1 == n)
+            ans = 1;
+    }
+    else
+    {
+        for (long long i = max(1LL, n - 50); i <= n; i++)
+            for (long long j = max(1LL, n - 50); j <= n; j++)
+                for (long long k = max(1LL, n - 50); k <= n; k++)
+                {
+                    ans = max(ans, lcm(i, lcm(j, k)));
+                }
+    }
+    cout << ans << "\n";
+    return 0;
+}
\ No newline at end of file