Merge branch 'main' into main

Author: Dheeraj-06

Problems/Mathematics/Day-01/q001.md

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+# PROBLEM STATEMENT :
+
+There are **n people** standing in a circle. They are numbered from **1 to n** in a clockwise order. The circle is **even** (i.e. *n* is even).
+
+Each person is looking at the person standing exactly opposite him in the circle.
+
+You are given three **distinct** integers **a**, **b**, and **c**. It is known that person **a** is looking at person **b**.
+
+Determine the number of the person that person **c** is looking at.  
+If there are **multiple answers**, print **any** of them.  
+If there is **no answer**, print **-1**.
+
+---
+
+## Input
+
+The first line contains a single integer **t**  
+(**1 ≤ t ≤ 10⁴**) — the number of test cases.
+
+Each of the following **t** lines contains three distinct integers **a**, **b**, and **c**  
+(**1 ≤ a, b, c ≤ 10⁸**).
+
+---
+
+## Output
+
+For each test case, print the answer —  
+the number of the person that person **c** is looking at,  
+or **-1** if the answer does not exist.
+
+---
+
+## Example
+
+### Input
+```
+7
+6 2 4
+2 3 1
+2 4 10
+5 3 4
+1 3 2
+2 5 4
+4 3 2
+```
+
+
+### Output
+```
+8
+-1
+-1
+-1
+4
+1
+-1
+```
+The problem statement is given just in case the codeforces server lags.<br>
+Submit your solution in the link given below and send a PR upon receiving AC.
+
+[PROBLEM LINK](https://codeforces.com/problemset/problem/1560/B)
+
+

Problems/Mathematics/Day-01/q002.md

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+## Problem Description
+
+You are given an integer `n` and an integer `k` (`k ≥ 2`), along with an array of `n` positive integers  
+`a1, a2, ..., an`.
+
+Your task is to count the number of unordered index pairs `(i, j)` such that:
+
+- `1 ≤ i < j ≤ n`
+- The product `ai × aj` is a **perfect k-th power**
+
+A number is called a perfect k-th power if it can be written in the form `x^k` for some integer `x`.
+
+---
+
+## Input Format
+
+- The first line contains two integers `n` and `k`.
+- The second line contains `n` integers `a1, a2, ..., an`.
+
+---
+
+## Output Format
+
+- Print a single integer — the number of valid pairs `(i, j)`.
+
+---
+
+## Constraints
+
+- `2 ≤ n ≤ 100000`
+- `2 ≤ k ≤ 100`
+- `1 ≤ ai ≤ 100000`
+
+---
+
+## Sample Input
+
+```
+6 3
+1 3 9 8 24 1
+```
+---
+
+## Output
+
+```
+5
+```
+
+[PROBLEM LINK]( https://codeforces.com/problemset/problem/1225/D)

Problems/Mathematics/Day-01/q1.cpp

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+//adding submission link as suggested earlier
+//https://codeforces.com/problemset/submission/1560/355167404
+
+#include <bits/stdc++.h>
+using namespace std;
+ 
+int main() {
+     int t;
+    cin >> t;
+    while(t--) {
+        int a,b,c;
+        cin>>a>>b>>c;
+        if((abs(b-a))*2<max(a,max(b,c))){
+            cout << -1 <<'\n';
+            continue;
+        }
+        if(c+(abs(b-a))>(abs(b-a)*2)){
+            cout<<(c+(abs(b-a)))%(abs(b-a)*2)<<'\n';
+        }
+        else{
+            cout<<c+(abs(b-a))<<'\n';
+        }
+    }
+    
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Arnav_Singh/Solution1.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int t;
+    cin >> t;
+    while (t--) {
+        int a, b, c;
+        cin >> a >> b >> c;
+
+        int diff = abs(a - b);
+        int n = 2 * diff;
+        if (diff == 0 || a > n || b > n || c > n) {
+            cout << -1 << '\n';
+            continue;
+        }
+        int half = n / 2;
+        int d;
+        if (c + half <= n)
+            d = c + half;
+        else
+            d = c - half;
+        cout << d << '\n';
+    }
+
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Avaneesh Verma/solution001.py

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+// CODEFORCES SUBMISSION LINK: https://codeforces.com/contest/1560/submission/355168755
+
+/*
+PROBLEM STATEMENT :
+
+There are n people standing in a circle. They are numbered from 1 to n in a clockwise order. The circle is even (i.e. n is even).
+
+Each person is looking at the person standing exactly opposite him in the circle.
+
+You are given three distinct integers a, b, and c. It is known that person a is looking at person b.
+
+Determine the number of the person that person c is looking at.
+If there are multiple answers, print any of them.
+If there is no answer, print -1.
+*/
+
+/*
+Approach:
+We first calculate the size of the circle. Knowing 'a' and 'b' is enough to know the size of the circle.
+Because we can know the number of people between 'a' and 'b' which would give us half the total number.
+We can easily get the person sitting opposite to any number once we know the size of the circle.
+*/
+
+/*
+Time Complexity: O(1)
+Space Complexity: O(1)
+*/
+
+import java.util.Scanner;
+
+public class Solution1 {
+    public static void main(String[] args) {
+        Scanner sc = new Scanner(System.in);
+        int t = sc.nextInt();
+        while (t-- > 0) {
+            int a = sc.nextInt();
+            int b = sc.nextInt();
+            int c = sc.nextInt();
+            int d;
+
+            int n = 2 * Math.abs(a - b);
+            if (a > n || b > n || c > n)
+                d = -1;
+            else {
+                int res = (c + (n / 2)) % n;
+                d = res != 0? res : n;
+            }
+
+            System.out.println(d);
+        }
+        sc.close();
+    }
+}

Problems/Mathematics/Day-01/sol/Ayush_Saha/Solution1.class

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+
+//Time: O(t)
+
+//Space: O(1)
+
+#include <bits/stdc++.h>
+using namespace std;
+int main(){
+  int t;
+    cin >> t;
+while(t--){
+ long long a, b, c;
+        cin >>  a >> b >> c;
+long long dist=llabs(a - b);
+long long n=2*dist;
+if(dist==0 || c>n){
+cout << -1 << '\n';
+continue;
+}
+long long half=n/2;
+if(c+half<=n){
+  cout << c+half << '\n';
+}
+else{
+  cout << c-half << '\n';
+
+}
+}
+ return 0;
+}
+

Problems/Mathematics/Day-01/sol/Ayush_Saha/Solution1.java

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+#include <bits/stdc++.h>
+using namespace std;
+
+#define ll long long
+#define pb push_back
+#define f(i,a,b) for (ll i = a; i < b; i++)
+#define fb(i,a,b) for (ll i = a; i >= b; i--)
+#define all(v) (v).begin(), (v).end()
+
+typedef vector<int> vi;
+typedef vector<ll> vl;
+typedef pair<int,int> pii;
+typedef pair<ll,ll> pll;
+
+#define fastio ios::sync_with_stdio(false); cin.tie(NULL);
+
+ll binexp(ll a, ll b) {
+    ll res = 1;
+    while (b > 0) {
+        if (b & 1) res = res * a;
+        a = a * a;
+        b >>= 1;
+    }
+    return res;
+}
+
+void solve() {
+    int a,b,c;
+    cin>>a>>b>>c;
+    if(c>2*abs(a-b) or a>2*abs(a-b) or b>2*(abs(a-b))){
+        cout<<-1<<endl;
+    }
+    else{
+        if(c>abs(a-b)) cout<<c-abs(a-b)<<endl;
+        else cout<<c+abs(a-b)<<endl;
+    }
+}
+
+int main() {
+    fastio;
+    int t = 1;
+    cin >> t;
+    while (t--) solve();
+}

Problems/Mathematics/Day-01/sol/BEESA_MANISH/Solution1.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+#define ll long long
+#define pb push_back
+#define f(i,a,b) for(ll i=a;i<b;i++)
+#define fb(i,a,b) for(ll i=a;i>=b;i--)
+#define vi vector<int>
+#define vl vector<ll>
+#define pii pair<int,int>
+#define pll pair<ll,ll>
+#define vpii vector<pii>
+#define vpll vector<pll>
+#define all(x) (x).begin(), (x).end()
+#define fast_io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
+
+const ll MOD = 1e9+7;
+const ll INF = 1e18;
+/*
+We need to count the no of pairs such that their product makes some number's k-th power-this means the product of the two numbers should have
+factors frequency that is divisible by k. For every number first factorize it using spf(also store its factors mod k) and also store how much
+factor more it needs to make the k-th power add this for all numbers.
+
+*/
+
+/* Submission Link: https://codeforces.com/contest/1225/submission/355183203 */
+void solve() {
+    int n,k;
+    cin>>n>>k;
+    vi a(n);
+    f(i,0,n) cin>>a[i];
+    vi spf(1e5+1);
+    f(i,1,1e5+1) spf[i]=i;
+    for(int i=2;i*i<=1e5;i++){
+        if(spf[i]==i){
+            for(int j=i*i;j<=1e5;j+=i){
+                if(spf[j]==j){
+                    spf[j]=i;
+                }
+            }
+        }
+    }
+    map<vpii,ll> cnt;
+    ll ans=0;
+    for(int x:a){
+        map<int,int> factors;
+        while(x>1){
+            int p=spf[x];
+            int c=0;
+            while(x%p==0){
+                x/=p;
+                c++;
+            }
+            c%=k;
+            if(c>0)
+                factors[p]=c;
+        }
+        vpii key,need;
+        for(auto &it:factors){
+            key.pb(it);
+            int y=(k-it.second)%k;
+            if(y>0){
+                need.pb({it.first,y});
+            }
+        }
+        ans+=cnt[need];
+        cnt[key]++;
+    }
+    cout<<ans<<endl;
+}
+
+int main() {
+    fast_io;
+    int t = 1;
+    //cin >> t;
+    while(t--) solve();
+    return 0;
+}