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RajwardhanShindeRajwardhanShinde
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Added some concepts and problems
1 parent 6548240 commit ab88eff

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‎Number Theory/(ATCODER)177E.cpp

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#include<bits/stdc++.h>
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#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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#define endl "\n"
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#define int long long int
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using namespace std;
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int mod = 1e9 + 7;
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bool ok = false;
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int fact[1000001], cnt[1000001];
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// This factors can be optimized using prime sieve method
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// void factors(int x) {
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// if (ok)
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// return;
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// for (int i = 1; i * i <= x; i++) {
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// if (x % i == 0) {
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// int a = i, b = x / i;
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// if (a == b) {
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// a = i;
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// b = 1;
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// }
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// if (a != 1) {
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// if (fact[a]) {
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// ok = true;
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// return;
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// }
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// fact[a] = true;
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// }
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// if (b != 1) {
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// if (fact[b]) {
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// ok = true;
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// return;
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// }
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// fact[b] = true;
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// }
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// }
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// }
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// }
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void sieve() {
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int n = 1000000;
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for (int i = 1; i <= n; i++)
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fact[i] = i;
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for (int i = 2; i <= n; i++) {
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if (fact[i] == i) {
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for (int j = i; j <= n; j += i)
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fact[j] = i;
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}
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}
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}
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int32_t main() {
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#ifndef ONLINE_JUDGE
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freopen("input.txt", "r", stdin);
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freopen("output.txt", "w", stdout);
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#endif
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fastio;
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sieve();
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int n;
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cin >> n;
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int arr[n], maxx = 0;
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int gcd = 0;
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for (int i = 0; i < n; i++) {
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cin >> arr[i];
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maxx = max(arr[i], maxx);
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gcd = __gcd(gcd, arr[i]);
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}
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bool flag = false;
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for (int i = 0; i < n; i++) {
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set<int> cur;
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while (arr[i] != 1) {
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cur.insert(fact[arr[i]]);
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arr[i] /= fact[arr[i]];
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}
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for (auto c : cur) {
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cnt[c]++;
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if (cnt[c] > 1) {
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flag = true;
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break;
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}
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}
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if (flag)
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break;
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}
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if (!flag)
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cout << "pairwise coprime";
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else if (gcd == 1)
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cout << "setwise coprime";
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else
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cout << "not coprime";
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return 0;
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}

‎Number Theory/(CODECHEF)GCDQUERIES.cpp

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#include<bits/stdc++.h>
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#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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#define endl "\n"
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#define int long long int
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using namespace std;
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#define N 101
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int arr[100001], pref[100001], suff[100001];
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// https://www.codechef.com/problems/GCDQ
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// Exclude L and R part from gcd and calculate gcd for rest array
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// Idea is to prepare prefix and suffix array of gcds and take gcd of pref[l - 1] and suff[r + 1]
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int mod = 1e9 + 7;
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int gcd(int a, int b) {
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while(b) {
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int temp = b;
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b = a % b;
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a = temp;
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}
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return a;
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}
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int32_t main() {
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int t, q, l, r, n;
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cin >> t;
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while(t--) {
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cin >> n >> q;
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for(int i = 1; i <= n; i++)
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cin >> arr[i];
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pref[0] = 0;
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for(int i = 1; i <= n; i++)
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pref[i] = gcd(pref[i - 1], arr[i]);
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suff[n + 1] = 0;
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for(int i = n; i >= 1; i--)
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suff[i] = gcd(suff[i + 1], arr[i]);
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while(q--) {
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cin >> l >> r;
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cout << gcd(pref[l - 1], suff[r + 1]) << endl;
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}
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}
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return 0;
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}

‎Number Theory/(CODECHEF)ModularGCD.cpp

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#include<bits/stdc++.h>
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#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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#define endl "\n"
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#define int long long int
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using namespace std;
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int mod = 1e9 + 7;
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// https://www.codechef.com/problems/GCDMOD
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// As here we have to find gcd of a ^ n + b ^ n and |a - b| so its good to find candidates for second than first as a - b would have less values to check
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// calculating a ^ n + b ^ n can be huge but here we are using the number we got from candidates of a - b as mod for a ^ n and b ^ n that would reduce the calculations
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int power(int a, int n, int m) {
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int res = 1;
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while(n) {
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if(n & 1) {
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res = ((res % m) * (a % m)) % m;
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n--;
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} else {
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a = ((a % m) * (a % m)) % m;
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n /= 2;
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}
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}
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return res;
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}
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int gcd(int a, int b, int n) {
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// a == b then a - b == 0 so we are adding power of a and b
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if(a == b)
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return (power(a, n, mod) + power(b, n, mod)) % mod;
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// here we will find candidates of a - b till sqrt(a - b) and we will choose maximum candidate that divides a - b
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int num = a - b;
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int candidate = 1;
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for(int i = 1; i * i <= num; i++) {
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if(num % i == 0) {
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int temp = (power(a, n, i) + power(b, n, i)) % i;
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if(temp == 0)
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candidate = max(candidate, i);
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temp = (power(a, n, num / i) + power(b, n, num / i)) % (num / i);
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if(temp == 0)
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candidate = max(candidate, num / i);
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}
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}
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return candidate % mod;
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}
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int32_t main() {
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int t, a, b, n;
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cin >> t;
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while(t--) {
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cin >> a >> b >> n;
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cout << gcd(a, b, n) << endl;
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}
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return 0;
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}

‎Number Theory/(CODEFORCES)742A.cpp

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#include<bits/stdc++.h>
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#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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#define endl "\n"
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#define int long long int
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using namespace std;
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int power(int a, int n, int p) {
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int res = 1;
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while(n) {
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if(n & 1) {
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res = (res * a) % p;
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n--;
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} else {
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a = (a * a) % p;
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n /= 2;
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}
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}
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return res;
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}
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int32_t main() {
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int n;
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cin >> n;
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// 8 is congruence to 1378
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cout << power(8, n, 10);
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return 0;
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}

‎Number Theory/(HACKERRANK)FibonacciFinding.cpp

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#include<bits/stdc++.h>
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#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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#define endl "\n"
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#define int long long int
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using namespace std;
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#define N 3
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int T[N][N], I[N][N], arr[N];
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// https://www.hackerrank.com/challenges/fibonacci-finding-easy/problem
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int mod = 1e9 + 7;
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void mul(int A[][N], int B[][N], int dim) {
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int res[dim + 1][dim + 1];
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for(int i = 1; i <= dim; i++) {
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for(int j = 1; j <= dim; j++) {
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res[i][j] = 0;
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for(int k = 1; k <= dim; k++)
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res[i][j] = (res[i][j] + A[i][k] * B[k][j]) % mod;
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}
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}
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for(int i = 1; i <= dim; i++)
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for(int j = 1; j <= dim; j++)
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A[i][j] = res[i][j];
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}
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int getFib(int n) {
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if(n <= 2)
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return arr[n];
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// Identity Matrix
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I[1][1] = I[2][2] = 1;
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I[1][2] = I[2][1] = 0;
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// Transition Matrix for sequence
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T[1][1] = 0;
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T[1][2] = T[2][1] = T[2][2] = 1;
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// as transition matrix raise to n - 1 for nth term
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n--;
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while(n) {
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if(n & 1)
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mul(I, T, 2), n--;
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else
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mul(T, T, 2), n /= 2;
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}
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return (arr[1] * I[1][1] + arr[2] * I[2][1]) % mod;
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}
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int32_t main() {
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int t, n;
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cin >> t;
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while(t--) {
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cin >> arr[1] >> arr[2] >> n;
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n++;
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cout << getFib(n) << endl;
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}
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return 0;
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}
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‎Number Theory/BinomialCoefficient.cpp

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#include<bits/stdc++.h>
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#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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#define endl "\n"
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#define int long long int
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using namespace std;
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int fact[1000001];
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int mod = 1e9 + 7;
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// Calculate C(n, k) % mod where mod > n
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// formula is n! / k! * (n - k)! but here we are using mod and precalculating factorial
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int power(int a, int n) {
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int res = 1;
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while(n) {
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if(n & 1)
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res = (res * a) % mod, n--;
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else
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a = (a * a) % mod, n /= 2;
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}
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return res;
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}
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int C(int n, int k) {
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if(k > n)
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return 0;
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int res = fact[n];
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// here instead of dividing we use inverse
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res = (res * power(fact[k], mod - 2)) % mod;
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res = (res * power(fact[n - k], mod - 2)) % mod;
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return res;
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}
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int32_t main() {
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int t, n, k;
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cin >> t;
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fact[0] = fact[1] = 1;
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for(int i = 2; i <= 1000000; i++)
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fact[i] = (fact[i - 1] * i) % mod;
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while(t--) {
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cin >> n >> k;
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cout << C(n, k) << endl;
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}
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return 0;
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}

‎Number Theory/FermatsInverse.cpp

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#include<bits/stdc++.h>
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#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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#define endl "\n"
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#define int long long int
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using namespace std;
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int power(int a, int n, int m) {
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int res = 1;
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while(n) {
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if(n & 1) {
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res = (res * a) % m;
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n--;
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} else {
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a = (a * a) % m;
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n /= 2;
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}
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}
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return res;
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}
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int32_t main() {
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int t, a, m;
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cin >> t;
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while(t--) {
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cin >> a >> m;
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cout << power(a, m - 2, m) << endl;
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}
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return 0;
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}

‎Number Theory/PrimeFactorization.cpp

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#include<bits/stdc++.h>
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#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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#define endl "\n"
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#define int long long int
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using namespace std;
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// Complexity - O(sqrt(N))
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void primeFact(int n) {
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for(int i = 2; i * i <= n; i++) {
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if(n % i == 0) {
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int cnt = 0;
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while(n % i == 0) {
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cnt++;
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n /= i;
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}
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cout << i << "^" << cnt << endl;
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}
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}
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// If number is prime then it would need itself raise to 1 as no number would divide it
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if(n > 1)
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cout << n << "^" << 1 << endl;
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}
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int32_t main() {
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int n;
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cin >> n;
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primeFact(n);
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return 0;
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}

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