Merge branch 'opencodeiiita:main' into main

Author: Rushalverma

Problems/Mathematics/Day-01/sol/Adarsh Yadav/solution.cpp

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+//https://codeforces.com/contest/1560/problem/C
+//https://codeforces.com/contest/1560/submission/355210608
+// TC : O(1) SC: O(1)
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int t;
+	cin>>t;
+	while(t--){
+	    int a,b,c;
+	    cin>>a>>b>>c;
+	    int k=abs(a-b);
+	    if(a>2*k || b>2*k || c>2*k){
+	        cout<<-1<<endl;
+	    }
+	    else{
+	        if((c-k)>0){
+	            cout<<c-k<<endl;
+	        }
+	        else{
+	            cout<<c+k<<endl;
+	        }
+	    }
+	}
+
+}

Problems/Mathematics/Day-01/sol/Aditi_Guin/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 half = abs(a - b);
+        int n = 2 * half;
+
+        if (c > n) {
+            cout << -1 << "\n";
+        } else {
+            int opposite = (c + half <= n) ? (c + half) : (c - half);
+            cout << opposite << "\n";
+        }
+    }
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Aditi_Guin/Solution2.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, k;
+    cin >> n >> k;
+
+    vector<int> a(n);
+    for (int &x : a) cin >> x;
+
+    map<vector<pair<int,int>>, long long> freq;
+    long long ans = 0;
+
+    for (int x : a) {
+        int temp = x;
+        vector<pair<int,int>> cur, need;
+
+        for (int p = 2; p * p <= temp; p++) {
+            if (temp % p == 0) {
+                int cnt = 0;
+                while (temp % p == 0) {
+                    temp /= p;
+                    cnt++;
+                }
+                cnt %= k;
+                if (cnt) {
+                    cur.push_back({p, cnt});
+                    need.push_back({p, (k - cnt) % k});
+                }
+            }
+        }
+
+        if (temp > 1) {
+            cur.push_back({temp, 1 % k});
+            need.push_back({temp, (k - 1) % k});
+        }
+
+        ans += freq[need];
+        freq[cur]++;
+    }
+
+    cout << ans << "\n";
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Aditya_Singh/solution002.cpp

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+/*
+Problem:
+You are given an array of n positive integers and an integer k (k >= 2).
+Count the number of unordered pairs (i, j) such that ai * aj is a perfect k-th power.
+
+Approach:
+1. Precompute smallest prime factor (SPF) up to max(ai).
+2. For each number, compute its prime factorization.
+3. Reduce exponents modulo k to form a "signature".
+4. Compute the complementary signature needed to make exponents divisible by k.
+5. Use a map to count matching signatures seen so far.
+
+Time Complexity: O(n log A)
+Space Complexity: O(n)
+
+Problem Link:
+https://codeforces.com/problemset/problem/1225/D
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+static const int MAXA = 100000;
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, k;
+    cin >> n >> k;
+
+    vector<int> a(n);
+    for (int i = 0; i < n; i++) {
+        cin >> a[i];
+    }
+
+    // Step 1: Compute Smallest Prime Factor (SPF)
+    vector<int> spf(MAXA + 1);
+    for (int i = 1; i <= MAXA; i++) spf[i] = i;
+
+    for (int i = 2; i * i <= MAXA; i++) {
+        if (spf[i] == i) {
+            for (int j = i * i; j <= MAXA; j += i) {
+                if (spf[j] == j)
+                    spf[j] = i;
+            }
+        }
+    }
+
+    // Map to store frequency of signatures
+    map<vector<pair<int,int>>, long long> freq;
+    long long answer = 0;
+
+    // Step 2–5: Process each number
+    for (int x : a) {
+        map<int,int> factorCount;
+
+        // Factorize using SPF
+        while (x > 1) {
+            int p = spf[x];
+            factorCount[p]++;
+            x /= p;
+        }
+
+        vector<pair<int,int>> signature;
+        vector<pair<int,int>> complement;
+
+        for (auto &it : factorCount) {
+            int prime = it.first;
+            int exp = it.second % k;
+
+            if (exp != 0) {
+                signature.push_back({prime, exp});
+                complement.push_back({prime, (k - exp) % k});
+            }
+        }
+
+        // Count valid pairs
+        answer += freq[complement];
+
+        // Store current signature
+        freq[signature]++;
+    }
+
+    cout << answer << "\n";
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Avaneesh Verma/solution002.py

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+#include <bits/stdc++.h>
+using namespace std;
+ 
+int main() {
+    int n, k;
+    cin >> n >> k;
+ 
+    vector<int> a(n);
+    for (int i = 0; i < n; i++)
+        cin >> a[i];
+ 
+    map<vector<pair<int,int>>, long long> freq;
+    long long ans = 0;
+ 
+    for (int i = 0; i < n; i++) {
+        int x = a[i];
+        vector<pair<int,int>> cur, need;
+ 
+        for (int p = 2; p * p <= x; p++) {
+            if (x % p == 0) {
+                int cnt = 0;
+                while (x % p == 0) {
+                    x /= p;
+                    cnt++;
+                }
+                cnt %= k;
+                if (cnt > 0) {
+                    cur.push_back({p, cnt});
+                    need.push_back({p, (k - cnt) % k});
+                }
+            }
+        }
+ 
+        if (x > 1) {
+            cur.push_back({x, 1});
+            need.push_back({x, (k - 1) % k});
+        }
+        sort(cur.begin(), cur.end());
+        sort(need.begin(), need.end());
+ 
+        ans += freq[need];
+        freq[cur]++;
+    }
+    cout << ans << "\n";
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Ayush Mishra/Solution1.cpp

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+// submission link
+// https://codeforces.com/contest/1560/submission/355180160
+
+// Space Complexity = O(1)
+// Time Complexity = O(1) for main solution code O(t) for overall code
+
+
+
+// The if condition checks if the circle is valid and if the inputs exist within that circle.
+// abs(b-a)<=1: Ensures a and b are not adjacent or the same.
+// 2*(abs(b-a)-1)+2 < c || 2*(abs(b-a)-1)+2 < a || 2*(abs(b-a)-1)+2 < b.  checks that the input numbers a, b, and c are not larger than the total size of the circle.
+// If any of these are true, it prints -1 .
+
+// The line int k = 2*(abs(b-a)-1) + 2 calculates the total number of items in the circle.
+
+// If the inputs are valid, it calculates the number opposite to c.
+// The halfway distance is k / 2.
+// If c is in the first half (c<= k/2), the opposite is c + k/2.
+// If c is in the second half (c > k/2), the opposite is c - k/2.
+
+
+
+
+#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)<=1 || 2*(abs(b-a)-1)+2 < c || 2*(abs(b-a)-1)+2 < a || 2*(abs(b-a)-1)+2 < b ){
+            cout<<-1<<"\n";
+        }
+        else {
+            int k =  2*(abs(b-a)-1) + 2;
+            if(c<=k/2) cout<<c+k/2 <<"\n";
+            else cout<<c-k/2<<"\n";
+        }
+
+    }
+    return 0;
+}

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

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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.
+
+Approach : Total(i.e. n) = 2*(difference of a and b), so if any of the a,b,c is greater than total, it will return -1. Else we should return c + |a-b| or c-|a-b| (only one of them will be feasible)
+
+Time Complexity: O(1) because we have created a formula directly.
+Space Complexity : O(1) because no. of inputs is fixed 
+
+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 
+
+Submission link : https://codeforces.com/contest/1560/submission/355191015
+*/
+
+#include<bits/stdc++.h>
+#include<array>
+#include<unordered_set>
+using namespace std;
+typedef long long ll;
+#define int            long long int
+#define F              first
+#define S              second
+#define pb             push_back
+#define si             set <int>
+#define vi             vector <int>
+#define pii            pair <int, int>
+#define vpi            vector <pii>
+#define vpp            vector <pair<int, pii>>
+#define mii            map <int, int>
+#define mpi            map <pii, int>
+#define spi            set <pii>
+#define endl           "\n"
+#define sz(x)          ((int) x.size())
+#define all(p)         p.begin(), p.end()
+#define double         long double
+#define que_max        priority_queue <int>
+#define que_min        priority_queue <int, vi, greater<int>>
+#define bug(...)       __f (#__VA_ARGS__, __VA_ARGS__)
+#define vin(a) for(int i = 0; i< a.size();i++) cin>>a[i]
+#define vout(a) {for(auto x : a) cout << x << " "; cout << endl;}
+#define pp(a) for(auto x : a) cout << x.F << " " << x.S << endl
+#define print(a,x,y) {for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endl}
+
+inline int power(int a, int b){int x = 1;while (b){if (b & 1) x *= a;a *= a;b >>= 1;}return x;}
+template <typename Arg1>
+void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
+template <typename Arg1, typename... Args>
+void __f (const char* names, Arg1&& arg1, Args&&... args)
+{
+    const char* comma = strchr (names + 1, ',');
+    cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
+}
+
+bool isPrime(int n){ if(n <= 1) return false; for(int i=2;i*i<=n;i++) if(n % i == 0) return false; return true; }
+ll gcd(ll a, ll b){ return b ? gcd(b, a % b) : a; }
+ll lcm(ll a, ll b){ return a / gcd(a,b) * b; }
+ll modpow(ll a, ll b, ll m){ ll r=1; while(b){ if(b&1) r=r*a%m; a=a*a%m; b>>=1;} return r; }
+
+bool getBit(ll x, int k){ return (x >> k) & 1; }
+ll setBit(ll x, int k){ return x | (1LL << k); }
+ll clearBit(ll x, int k){ return x & ~(1LL << k); }
+ll toggleBit(ll x, int k){ return x ^ (1LL << k); }
+int countBits(ll x){ return __builtin_popcountll(x); }
+bool isPowerOfTwo(ll x){ return x && !(x & (x - 1)); }
+
+vector<int> sieve(int n){ vector<int> p(n+1,1), primes; p[0]=p[1]=0; for(int i=2;i*i<=n;i++) if(p[i]) for(int j=i*i;j<=n;j+=i) p[j]=0; for(int i=2;i<=n;i++) if(p[i]) primes.pb(i); return primes; }
+
+const int N = 200005;
+
+void solve() {
+    int a,b,c;
+    cin >> a >> b >> c;
+
+    int total = 2*abs(a-b);
+    if (max(max(a,b),c)>total)
+    {
+      cout << -1 << endl;return;
+    }
+    else
+    {
+      int ans = ((c-abs(a-b))>0 )? c-abs(a-b) : c+ abs(a-b) ;
+      cout << ans << endl;return;
+    }
+}
+
+int32_t main()
+{
+    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+    int t = 1;
+    cin >> t;
+    while (t--) solve();
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Satwik_Santosh/.cph/.Solution1.cpp_c67c6042fed1bb34ccbfc5368abbf7ee.prob

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+{"name":"Local: Solution1","url":"c:\\Project\\OpenCode25\\CP-Chronicles\\Problems\\Mathematics\\Day-01\\sol\\Satwik_Santosh\\Solution1.cpp","tests":[{"id":1766771594036,"input":"","output":""}],"interactive":false,"memoryLimit":1024,"timeLimit":3000,"srcPath":"c:\\Project\\OpenCode25\\CP-Chronicles\\Problems\\Mathematics\\Day-01\\sol\\Satwik_Santosh\\Solution1.cpp","group":"local","local":true}