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

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

‎Problems/Mathematics/Day-01/sol/Ibrahim Raza Beg/Solution1.cpp‎

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//https://codeforces.com/contest/1560/submission/355167905
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#include <bits/stdc++.h>
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using namespace std;
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#define int long long
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#define ll long long
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#define ull unsigned long long
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#define ld long double
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#define pii pair<int,int>
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#define vi vector<int>
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#define vvi vector<vector<int>>
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#define vpi vector<pair<int,int>>
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#define all(x) (x).begin(), (x).end()
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const long long MAXN = 2e6;
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const long long MOD = 1e9 + 7;
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const int INF = 1e18;
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const ld EPS = 1e-9;
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#define fastio ios::sync_with_stdio(false); cin.tie(nullptr);
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#ifndef ONLINE_JUDGE
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#define debug(x) cerr << #x << " = " << x << "\n";
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#else
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#define debug(x)
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#endif
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// Utility functions
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int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
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int lcm(int a, int b) { return a / gcd(a, b) * b; }
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int mod_add(int a, int b, int m=MOD) { return ((a % m) + (b % m) + m) % m; }
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int mod_sub(int a, int b, int m=MOD) { return ((a % m) - (b % m) + m) % m; }
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int mod_mul(int a, int b, int m=MOD) { return ((a % m) * (b % m)) % m; }
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//Binary Exponentiation
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int binexp(int a, int b, int m=MOD) {
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int res = 1;
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a %= m;
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while(b > 0) {
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if(b & 1) res = (res * a) % m;
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a = (a * a) % m;
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b >>= 1;
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}
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return res;
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}
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int mod_inv(int a, int m=MOD) {
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return binexp(a, m - 2, m);
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}
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long long fac[MAXN + 1];
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long long inv[MAXN + 1];
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long long exp(long long x, long long n, long long m) {
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x %= m;
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long long res = 1;
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while (n > 0) {
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if (n % 2 == 1) { res = res * x % m; }
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x = x * x % m;
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n /= 2;
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}
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return res;
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}
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void factorial() {
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fac[0] = 1;
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for (long long i = 1; i <= MAXN; i++) { fac[i] = fac[i - 1] * i % MOD; }
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}
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void inverses() {
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inv[MAXN] = exp(fac[MAXN], MOD - 2, MOD);
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for (long long i = MAXN; i >= 1; i--) { inv[i - 1] = inv[i] * i % MOD; }
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}
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long long choose(long long n, long long r) {
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if (r > n)return 0ll;
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return (fac[n] * inv[r] % MOD * inv[n - r] % MOD) % MOD;
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}
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long long catalan(long long n) {
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return (exp(n + 1, MOD - 2, MOD) % MOD * choose(2 * n, n) % MOD) % MOD;
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}
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//Returns all prime numbers <=N
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vector<int> sieve(int n) {
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vector<bool> prime(n + 1, true);
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for (int p=2;p*p<=n;p++) {
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if (prime[p] == true) {
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for (int i=p*p;i<=n;i+=p)
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prime[i] = false;
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}
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}
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vector<int> res;
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for (int p = 2; p <= n; p++){
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if (prime[p]){
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res.push_back(p);
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}
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}
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return res;
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}
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//Prime factors of all numbers upto N
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vector<vector<int>> primefactors(int N)
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{
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vvi pfac(N + 1);
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for (int i=2;i<=N;i++){
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if (!pfac[i].empty())
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continue;
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for (int j = i; j <= N; j += i)
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pfac[j].push_back(i);
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}
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return pfac;
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}
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//smallest prime factor of a number
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int spf(int n) {
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if (n % 2 == 0) return 2;
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for (int i = 3; i * i <= n; i += 2) {
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if (n % i == 0) return i;
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}
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return n;
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}
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int power2(int p)
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{
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int v=1ll<<p;
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return v;
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}
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//Sort functions
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void sort(vector<int>&a)
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{
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sort(a.begin(),a.end());
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}
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void psort(vector<pair<int,int>>&a)
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{
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sort(a.begin(),a.end());
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}
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void rsort(vector<int>&a)
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{
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sort(a.rbegin(),a.rend());
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}
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void rpsort(vector<pair<int,int>>&a)
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{
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sort(a.rbegin(),a.rend());
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}
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//2-D Vector Declaration:
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//vvi mat(n,vi(m)); // nxm matrix, all initialized to 0
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/***************Code***************/
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void solve() {
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int a,b,c;
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cin>>a>>b>>c;
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int diff=abs(a-b);
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int p=2*diff;
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if(a<=p&&b<=p&&c<=p)
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{
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if(c>diff)
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cout<<c-diff<<endl;
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else
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cout<<c+diff<<endl;
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}
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else
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cout<<"-1"<<endl;
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}
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signed main() {
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fastio;
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int t = 1;
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cin >> t;
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while (t--) {
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solve();
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}
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return 0;
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}

‎Problems/Mathematics/Day-01/sol/Prashant_dwivedi/Solution2.java‎

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/*
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Problem Statement:
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You are given an integer n and an integer k (k ≥ 2), along with an array of n
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positive integers.
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Your task is to count the number of unordered index pairs (i, j) such that:
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1 ≤ i < j ≤ n
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The product ai × aj is a perfect k-th power.
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---------------------------------------
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Approach:
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A number is a perfect k-th power if, in its prime factorization,
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the exponent of every prime is divisible by k.
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For each number:
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1. Factorize it using smallest prime factor (SPF).
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2. Reduce each prime’s exponent modulo k.
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3. Store this reduced form as a "signature".
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For a pair (ai, aj) to form a perfect k-th power:
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- Their signatures must complement each other such that
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the sum of exponents (mod k) becomes 0 for every prime.
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---------------------------------------
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Time Complexity:
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O(n log a)
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Space Complexity:
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O(n)
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---------------------------------------
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*/
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/*---------Problem link--------------------
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https://codeforces.com/problemset/status?my=on
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*/
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import java.util.*;
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public class Solution2 {
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static int MAX = 100000;
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static int[] spf = new int[MAX + 1];
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// Precompute smallest prime factor for every number
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static void computeSPF() {
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for (int i = 1; i <= MAX; i++)
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spf[i] = i;
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for (int i = 2; i * i <= MAX; i++) {
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if (spf[i] == i) {
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for (int j = i * i; j <= MAX; j += i) {
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if (spf[j] == j)
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spf[j] = i;
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}
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}
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}
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}
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// Build reduced prime-exponent signature
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static Map<Integer, Integer> buildSignature(int x, int k) {
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Map<Integer, Integer> map = new HashMap<>();
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while (x > 1) {
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int p = spf[x];
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int cnt = 0;
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while (x % p == 0) {
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x /= p;
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cnt++;
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}
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cnt %= k;
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if (cnt > 0)
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map.put(p, cnt);
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}
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return map;
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}
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// Build complement signature
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static Map<Integer, Integer> buildComplement(Map<Integer, Integer> sig, int k) {
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Map<Integer, Integer> comp = new HashMap<>();
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for (Map.Entry<Integer, Integer> e : sig.entrySet()) {
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comp.put(e.getKey(), (k - e.getValue()) % k);
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}
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return comp;
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}
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public static void main(String[] args) {
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Scanner sc = new Scanner(System.in);
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computeSPF();
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int n = sc.nextInt();
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int k = sc.nextInt();
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Map<Map<Integer, Integer>, Long> freq = new HashMap<>();
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long ans = 0;
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for (int i = 0; i < n; i++) {
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int x = sc.nextInt();
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Map<Integer, Integer> sig = buildSignature(x, k);
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Map<Integer, Integer> comp = buildComplement(sig, k);
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ans += freq.getOrDefault(comp, 0L);
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freq.put(sig, freq.getOrDefault(sig, 0L) + 1);
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}
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System.out.println(ans);
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sc.close();
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}
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}

‎contributers.md‎

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| Name | GitHub ID | College Name |
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|------------------------|-------------------------|------------------------------------------------------------------|
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| Pradip Maity | @codeboy-pro | Haldia Institute Of Technology |
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| Siddhant Shekhar | sshekhar563 | United University |
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| Ayush Saha | verifiedHuman18 | IIIT Allahabad |
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| Khushi Shorey | khushishorey | IIIT Allahabad |
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| Nitin Kumar | Nitinkr03 | IIIT Allahabad
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| Manisha katariya | chanchukatariya5 | IIIT Allahabad |
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| Abhigyan Tiwari | Abh-igyan | NIT Silchar |
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| Harsh Verma | harshverma14 | IIIT Allahabad |
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| Ayush Mishra | ayush-mg | IIIT Allahabad |
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| Ronak Goyal | ronakgoyal1 | IIIT Allahabad |
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| Sayed Al Amaan Zaidi | amaan1114 | Rishihood University |
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| Somesh Pawan Kamad | Entropy-rgb | IIIT Hyderabad |
42-
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| Roshan Gupta | roshangupta4804-hue | IIIT Allahabad
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<!-- ADD ABOVE THIS -->
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<!-- example | Korvac | Betty | Reyansh College | -->

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