Merge branch 'main' into main

Author: vraj826

Problems/Mathematics/Day-01/sol/Abhay/Q1sol.cpp

@@ -0,0 +1,53 @@
+#include <bits/stdc++.h>
+using namespace std;
+/*
+Logic:
+The main key in solving the problem is that a,b will be enought to know that the size of the area.
+Another observation would be that 1 is always facing the element  more than the element which faces 
+at the last element or the size of the circle.
+Once, you get the size of the array, we just have to check if the any of the elements are larger than 
+the size of the array. 
+On deeper checks like- a=3, b=2,c=1;
+we can observe how this isn't true but how does the conditon fulfill the case?
+--> the max check covers all small checks, like the circle don being possible
+in cases like a=1,b=6, c=100
+--> the circle condition is also checked by the max case. 
+
+So, after the checking, we need to see the answer, that can be easily given as the
+difference is same for all pairs of elements  modular nature of the circle( cool words that make sense)
+ 
+Time complexity-->
+
+The time complexity is O(1)
+The space complexity is O(1)
+
+Submission link- https://codeforces.com/contest/1560/submission/355188467
+*/
+//Soln->
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+
+    while (t--) {
+        int a,b,c;
+        cin>>a;
+        cin>>b;
+        cin>>c;
+        int x=abs(b-a);
+    if(max({a,b,c})>(2*x))
+    cout<<-1<<endl;
+    else
+    {
+        if(c>x)
+        cout<<c-x<<endl;
+        else
+        cout<<c+x<<endl;
+    }
+    
+    }
+    return 0;
+}
+    

Problems/Mathematics/Day-01/sol/Coldesy/Coldesy.cpp

@@ -0,0 +1,38 @@
+//https://codeforces.com/contest/1560/submission/355205808
+//time complexity:O(t)
+//space complexity:O(1)
+#include <iostream>
+using namespace std;
+
+int main()
+{
+    int t;
+    cin>>t;
+    while(t--){
+        int a,b,c;
+        cin>>a>>b>>c;
+        int len;
+        if(a>b){
+            len=a-b;
+            len=2*len;
+        }               // staring at opposite end means both of them divide the circle 
+                        // in exact half so we substract them and multiply by 2 to get exact length
+        else {
+           
+            len=b-a;
+            len=2*len;
+        
+        }
+        if(a<=len&&b<=len&&c<=len){//a,b,c ofcourse cannot be greater than length
+            if(c+(len/2)>len){
+                cout<<c-(len/2)<<endl;//c will either stare at +n/2 or -n/2
+            }
+            else{
+                cout<<c+(len/2)<<endl;
+            }
+        }
+        else{cout<<-1<<endl;}
+    }
+
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Kunwar Gaba/Solution1.cpp

@@ -0,0 +1,31 @@
+#include <bits/stdc++.h>
+using namespace std;
+#define int long long int
+#define double long double
+#define endl '\n'
+void resolve() {
+    int a,b,c,d=0;
+    cin>>a>>b>>c;
+    int n = 2*abs(b-a);
+    if(b>n || c>n || a>n)
+    d = -1;
+    else
+    {
+        if(c<=(n/2))
+        d = c + n/2;
+        else
+        d = c - n/2;
+
+        if(d==a || d==b || d>n || d<1)
+        d=-1;
+    }
+    cout<<d<<endl;
+}
+int32_t main() {
+	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
+	 int t;
+	 cin >> t;
+	 while(t--) 
+        resolve();
+	return 0;
+}

Problems/Mathematics/Day-02/q002.md

@@ -0,0 +1,73 @@
+## Problem Description
+
+Masha works in an advertising agency. In order to promote a new brand, she wants to conclude contracts with some bloggers.
+
+Masha has connections with `n` different bloggers. Blogger numbered `i` has `ai` followers.
+
+Due to a limited budget, Masha can only sign contracts with `k` different bloggers. She wants the advertisement to be seen by as many people as possible, so she must choose bloggers such that the **total number of followers is maximized**.
+
+Your task is to find the **number of ways** to select `k` bloggers so that the total number of their followers is maximum possible.
+
+Two ways are considered different if there is at least one blogger selected in one way but not in the other.
+
+---
+
+## Input Format
+
+- The first line contains an integer `t` — the number of test cases.
+- For each test case:
+  - The first line contains two integers `n` and `k`
+  - The second line contains `n` integers `a1, a2, ..., an`, where `ai` is the number of followers of the `i`-th blogger
+
+---
+
+## Output Format
+
+- For each test case, print a single integer — the number of ways to select `k` bloggers such that the total number of followers is maximum possible.
+- Since the answer can be large, print it modulo `10^9 + 7`.
+
+---
+
+
+
+## Constraints
+
+- `1 ≤ t ≤ 1000`
+- `1 ≤ k ≤ n ≤ 1000`
+- `1 ≤ ai ≤ n`
+- The sum of `n` over all test cases does not exceed `1000`
+
+---
+
+## Sample Input
+
+```
+3
+4 3
+1 3 1 2
+4 2
+1 1 1 1
+2 1
+1 2
+```
+---
+
+## Output
+
+```
+2
+6
+1
+```
+
+---
+
+### Additional Resource
+
+For understanding **modular arithmetic and modular inverses**, which are useful for solving problems with answers modulo `10^9+7`, you can check this video:  
+https://youtu.be/RCq5TYMZEwg
+
+---
+
+
+[PROBLEM LINK](http://codeforces.com/contest/1475/problem/E)

Problems/Mathematics/Day-02/q003.md

@@ -0,0 +1,60 @@
+## Problem Description
+
+You are given a single integer `N`.
+
+You are asked to compute the following summation:
+
+Σ(i=1 to N) Σ(j=1 to N) Σ(k=1 to i) Σ(l=1 to j) (gcd(k, i) × gcd(l, j) × gcd(i, j))
+
+Output the value of this summation **modulo 1000000007**.
+
+---
+
+## Input Format
+
+- The input consists of a single integer `N`.
+
+---
+
+## Output Format
+
+- Output the value of the summation modulo `1000000007`.
+
+---
+
+## Constraints
+
+- 1 ≤ N ≤ 10^6
+
+---
+
+
+## Sample Input 1
+
+```
+2
+```
+---
+
+## Output
+
+```
+25
+```
+---
+
+## Sample Input 2
+
+```
+8
+```
+---
+
+## Output
+
+```
+13348
+```
+---
+
+[PROBLEM LINK](https://www.codechef.com/problems/YETGCD?tab=statement)

contributers.md

@@ -45,10 +45,14 @@
 | Ronak Goyal            | ronakgoyal1             | IIIT Allahabad                                                   |
 | Sayed Al Amaan Zaidi           | amaan1114           | Rishihood University                                   |
 |Aditya Singh            | adityasingh-0803         | BVDUCOEP                                                        |
+| Ishan Raj Singh        | ishanrajsingh            | Amity, Noida     
+|Priyanshi Giri		| hellopaintinghi-cmd        | IIIT Allahabad                                               |
 | Mahavir dodiya         |Mvdodiya001               |  IIIT Allahabad                                                 |
 | Ishan Raj Singh        | ishanrajsingh            | Amity, Noida                                                    |
+| Harsh Garg             | gargharsh182005-lab      | IIIT Allahabad                                                  |
 | Lakshmish            | Coldesy             | IIIT Allahabad 
 |Priyanshu Ranjan        | justpriyanshu            | BIT Mesra
 | Vrajkumar Shah        | vraj826                   | DDU, Nadiad
+| Atharva Mehta          | Atharva-insane           |  IIIT Allahabad                                                  |                                               
 <!-- ADD ABOVE THIS -->
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